Maharashtra Board Class 12 Physics 2025 – Complete Solutions

Obtain / Derive an expression for average power dissipated in a series LCR circuit

Q: Is this compatible with MathJax?

Yes. All LaTeX expressions in this file work perfectly with MathJax.

Q: Can I use this in WordPress?

Yes. Paste this code inside a Custom HTML block.

Q: Is JavaScript required?

No. Everything works using only HTML and CSS.

Q: Is this SEO-friendly?

Yes. Semantic tags, headings, and structured content help SEO.

Published: December 15, 2025 •

Series LCR Circuit

1. Choose the correct option.

i) Which of the following phenomenon proves that light is a transverse wave?

(A) reflection
(B) interference
(C) diffraction
(D) polarization

Answer: (D) polarization

Explanation: Polarization occurs only for transverse waves. Light can be polarized, which proves it has transverse nature. Reflection, interference, and diffraction can occur for both transverse and longitudinal waves.

ii) Which property of light does not change when it travels from one medium to another?

(A) velocity
(B) wavelength
(C) amplitude
(D) frequency

Answer: (D) frequency

Explanation: When light passes from one medium to another, its velocity $v$ and wavelength $\lambda$ change, but its frequency $f$ remains constant because it depends on the source.

iii) When unpolarized light is passed through a polarizer, its intensity:

(A) increases
(B) decreases
(C) remains unchanged
(D) depends on the orientation of the polarizer

Answer: (B) decreases

Explanation: Unpolarized light contains waves in all directions. A polarizer only allows light oscillating in one direction to pass, reducing the intensity by half: \[ I = \frac{I_0}{2} \]

iv) In Young’s double slit experiment, the two coherent sources have different intensities. If the ratio of maximum intensity to the minimum intensity in the interference pattern produced is $25:1$. What was the ratio of intensities of the two sources?

(A) 5:1
(B) 25:1
(C) 3:2
(D) 9:4

Answer: (D) 9:4

Explanation: Let the intensities of the two sources be $I_1$ and $I_2$. For two sources with different amplitudes, the maximum and minimum intensities in interference pattern are given by: \[ I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2, \quad I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2 \] Given, $ \frac{I_{\max}}{I_{\min}} = 25 $ \[ \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = 25 \] \[ \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} = 5 \] \[ \sqrt{I_1} + \sqrt{I_2} = 5(\sqrt{I_1} - \sqrt{I_2}) \implies 6 \sqrt{I_2} = 4 \sqrt{I_1} \implies \frac{I_1}{I_2} = \frac{9}{4} \]

v) In Young’s double slit experiment, a thin uniform sheet of glass is kept in front of the two slits, parallel to the screen having the slits. The resulting interference pattern will:

(A) The interference pattern will remain unchanged
(B) The fringe width will decrease
(C) The fringe width will increase
(D) The fringes will shift

Answer: (D) The fringes will shift

Explanation: The thin glass introduces an additional path difference, causing a phase change. This shifts the positions of the bright and dark fringes. The fringe width $ \beta = \frac{\lambda D}{d} $ remains unchanged because it depends on wavelength, slit separation, and screen distance, which are not altered.

2. Answer in brief

(i) What are primary and secondary sources of light?

Answer:

Primary sources of light: Objects which emit light by themselves are called primary sources. Example: Sun, candle, electric bulb.
Secondary sources of light: Objects which do not emit light by themselves but reflect light from primary sources are called secondary sources. Example: Moon, planets, paper.

Answer:

A wavefront is the locus of all points in a medium which vibrate in the same phase.
Rays of light are always drawn perpendicular (normal) to the wavefront.
At a point far away from the source, the wavefront becomes a plane wavefront.

(iii) Why are multiple colours observed over a thin film of oil floating on water?

Answer:

Multiple colours are observed due to interference of light.

Light reflected from the upper and lower surfaces of the oil film interferes.
Different wavelengths satisfy conditions for constructive and destructive interference at different thicknesses.
Hence, different colours are seen at different points on the film.

(iv) In Young's double slit experiment, what will be observed when white light is incident but one slit has a red filter and the other a violet filter?

Answer:

No interference pattern will be observed.
Interference requires light waves of the same wavelength and coherent sources.
Red and violet lights have different wavelengths, hence the condition for interference is not satisfied.

(v) What is optical path length? How is it different from actual path length?

Answer:

Actual path length: The physical distance travelled by light in a medium.
Optical path length: The product of refractive index and actual path length.

$$ \text{Optical path length} = \mu \times \text{Actual path length} $$

Optical path length depends on the medium, whereas actual path length does not.

Q.3 Derive the laws of reflection of light using Huygens’ principle.

Answer:

Huygens’ Principle: Every point on a wavefront acts as a source of secondary wavelets, and the envelope of these wavelets gives the new wavefront.

Derivation:

Consider a plane wavefront (AB) incident on a reflecting surface (XY) as shown.

Point $A$ of the wavefront touches the reflecting surface first.
After time $t$, point $B$ reaches point $C$ on the surface.
In the same time, secondary wavelets from point $A$ spread with speed $v$ and reach point $D$.

Since: $$ AD = BC = vt $$ The reflected wavefront is the tangent (CD) drawn to the secondary wavelet. From geometry: $$ \angle i = \angle r $$ Also, the incident ray, reflected ray, and the normal lie in the same plane.

Thus, the laws of reflection are:

The angle of incidence equals the angle of reflection.
The incident ray, reflected ray, and normal lie in the same plane.

Q.4 Derive the laws of refraction of light using Huygens’ principle.

Answer:

Huygens’ Principle: Each point on a wavefront acts as a source of secondary wavelets, and the envelope of these wavelets forms the new wavefront.

Derivation:

Consider a plane wavefront (AB) incident obliquely on the boundary between two media with refractive indices (n_1) and (n_2).

Light travels with velocity $v_1$ in medium 1 and $v_2$ in medium 2.
Point $A$ reaches the boundary first and generates secondary wavelets.
In time $t$, point $B$ reaches point $C$.

Thus: $$ BC = v_1 t \quad \text{and} \quad AD = v_2 t $$ From geometry: $$ \sin i = \frac{BC}{AB}, \quad \sin r = \frac{AD}{AB} $$ Dividing: $$ \frac{\sin i}{\sin r} = \frac{v_1}{v_2} $$ But, $$ \frac{v_1}{v_2} = \frac{n_2}{n_1} $$ Hence: $$ \boxed{n_1 \sin i = n_2 \sin r} $$ This is **Snell’s law of refraction**.

Thus, the laws of refraction are:

The incident ray, refracted ray, and normal lie in the same plane.
The ratio $\dfrac{\sin i}{\sin r}$ is constant for a given pair of media.

Q.5 Explain what is meant by polarization and derive Malus’ law.

Answer:

Polarization:
Polarization is the phenomenon by which the vibrations of light waves are restricted to a single plane perpendicular to the direction of propagation. This proves that light is a transverse wave. Unpolarized light has vibrations in all possible planes perpendicular to the direction of propagation, whereas polarized light vibrates in only one plane.

Derivation of Malus’ Law:

Consider plane polarized light of intensity $I_0$ incident on an analyzer. Let $\theta$ be the angle between the plane of vibration of incident light and the transmission axis of the analyzer. The component of electric field along the transmission axis is: $$ E = E_0 \cos\theta $$ Since intensity is proportional to the square of the electric field: $$ I \propto E^2 $$ Therefore, $$ I = I_0 \cos^2\theta $$ This relation is known as Malus’ law.

$I$ = intensity of transmitted light
$I_0$ = intensity of incident polarized light
$\theta$ = angle between planes of polarization

Q.6 What is Brewster’s law? Derive the formula for Brewster angle.

Answer:

Brewster’s Law:
Brewster’s law states that when unpolarized light is incident on a transparent surface at a particular angle (called Brewster angle), the reflected light is completely plane polarized. The tangent of the Brewster angle is equal to the refractive index of the medium.

Derivation:

Let:

$i_B$ = Brewster angle
$r$ = angle of refraction
$\mu$ = refractive index of the medium

At Brewster angle: $$ i_B + r = 90^\circ $$ Using Snell’s law: $$ \mu = \frac{\sin i_B}{\sin r} $$ Since $r = 90^\circ - i_B$, $$ \sin r = \cos i_B $$ Substituting: $$ \mu = \frac{\sin i_B}{\cos i_B} $$ $$ \boxed{\mu = \tan i_B} $$ This is the required relation for Brewster’s angle.

Q.7 Describe Young’s double slit interference experiment and derive conditions for occurrence of dark and bright fringes on the screen. Define fringe width and derive a formula for it.

Answer:

Young’s Double Slit Experiment:
In Young’s double slit experiment, a monochromatic light source illuminates two narrow, closely spaced slits $S_1$ and $S_2$. These slits act as coherent sources and produce an interference pattern of alternate bright and dark fringes on a distant screen. The interference pattern is due to the superposition of light waves coming from the two slits.

Path Difference: For a point $P$ on the screen at a distance $y$ from the central maximum, the path difference is: $$ \Delta x = \frac{dy}{D} $$ where:

$d$ = distance between the slits
$D$ = distance of the screen from the slits

Condition for Bright Fringes: Bright fringes occur when the path difference is an integral multiple of wavelength. $$ \Delta x = n\lambda \quad (n = 0,1,2,\ldots) $$ $$ \frac{dy}{D} = n\lambda $$ $$ y_n = \frac{n\lambda D}{d} $$
Condition for Dark Fringes: Dark fringes occur when the path difference is an odd multiple of half wavelength. $$ \Delta x = (2n+1)\frac{\lambda}{2} $$ $$ y_n = \frac{(2n+1)\lambda D}{2d} $$
Fringe Width:
Fringe width ($\beta$) is defined as the distance between two consecutive bright fringes or two consecutive dark fringes.
Derivation of Fringe Width: Distance between two consecutive bright fringes: $$ \beta = y_{n+1} - y_n $$ $$ \beta = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} $$ $$ \boxed{\beta = \frac{\lambda D}{d}} $$

Q.8 What are the conditions for obtaining good interference pattern? Give reasons.

Answer:

The conditions for obtaining a clear and steady interference pattern are:

Coherent sources: The two sources must maintain a constant phase difference to produce sustained interference.
Monochromatic light: Light should have a single wavelength to obtain sharp and well-defined fringes.
Slits should be narrow: Narrow slits act as point sources and allow proper diffraction and overlap of waves.
Slits should be close to each other: Small separation increases fringe width, making fringes easily observable.
Screen should be far from slits: Large distance increases fringe separation and improves visibility.
Nearly equal intensities: Equal intensity waves produce maximum contrast between bright and dark fringes.

Q.9 What is meant by coherent sources? What are the two methods for obtaining coherent sources in the laboratory?

Answer:

Coherent Sources:
Two sources of light are said to be coherent if they emit light waves of the same frequency (or wavelength) and maintain a constant phase difference with time.

Methods of obtaining coherent sources:

Division of wavefront: A single wavefront is divided into two parts which act as coherent sources. Example: Young’s double slit experiment, Fresnel’s biprism.
Division of amplitude: The amplitude of a single beam is divided into two parts using reflection or refraction. Example: Interference using thin films, Newton’s rings.

Q.10 What is diffraction of light? How does it differ from interference? What are Fraunhofer and Fresnel diffractions?

Answer:

Diffraction of Light:
Diffraction is the phenomenon of bending and spreading of light waves when they pass through a narrow aperture or around an obstacle whose dimensions are comparable to the wavelength of light.

Difference between Interference and Diffraction:

Interference	Diffraction
Occurs due to superposition of light from two coherent sources	Occurs due to superposition of secondary wavelets from different parts of the same wavefront
Fringes are equally spaced	Fringes are not equally spaced
All bright fringes have equal intensity	Central maximum is brightest

Types of Diffraction:

Fraunhofer Diffraction: Occurs when the source and screen are effectively at infinite distance from the diffracting aperture. Parallel rays are used. It is observed using lenses.
Fresnel Diffraction: Occurs when the source and screen are at finite distances from the aperture. Wavefronts are spherical.

Q.11 Derive the conditions for bright and dark fringes produced due to diffraction by a single slit.

Answer:

Consider a single slit of width a illuminated by monochromatic light of wavelength λ. The diffraction pattern is observed on a distant screen. Let θ be the angle of diffraction. --- Condition for Dark Fringes (Minima):

For destructive interference, the path difference between rays from the top and bottom of the slit is: $$ a\sin\theta = n\lambda $$ where:

$n = 1, 2, 3, \dots$

Thus, the condition for dark fringes is: $$ \boxed{a\sin\theta = n\lambda} $$ The central maximum corresponds to $n = 0$. --- Condition for Bright Fringes (Maxima):

Bright fringes occur approximately midway between two successive minima. The condition for secondary maxima is given by: $$ a\sin\theta = \left(n + \frac{1}{2}\right)\lambda $$ where:

$n = 1, 2, 3, \dots$

The central bright fringe is the brightest and widest. --- Conclusion:

Central maximum is the brightest.
Intensity of secondary maxima decreases on either side.

Q.12 Describe Rayleigh’s criterion for resolution. Explain it for a telescope and a microscope.

Answer:

Rayleigh’s Criterion:

According to Rayleigh’s criterion, two point objects are said to be just resolved when the principal maximum of the diffraction pattern of one coincides with the first minimum of the other. --- Rayleigh’s Criterion for a Telescope:

For a telescope with objective aperture of diameter D, the minimum angular separation θ between two distant objects that can be just resolved is: $$ \boxed{\theta = 1.22 \frac{\lambda}{D}} $$ where:

$\lambda$ = wavelength of light used
$D$ = diameter of objective lens

Thus, resolving power increases with larger aperture. --- Rayleigh’s Criterion for a Microscope:

For a microscope, the minimum distance d between two points that can be just resolved is: $$ \boxed{d = 0.61 \frac{\lambda}{\mu \sin\theta}} $$ where:

$\mu$ = refractive index of medium
$\theta$ = half-angle of cone of light
$\mu \sin\theta$ = numerical aperture

A microscope has higher resolving power when numerical aperture is large. --- Conclusion:

Telescope resolution depends on aperture size.
Microscope resolution depends on numerical aperture.

Q.13 White light consists of wavelengths from 400 nm to 700 nm. What will be the wavelength range seen when white light is passed through glass of refractive index 1.55?

Answer:

When light enters a medium, its frequency remains constant but its wavelength changes. Relation between wavelength in vacuum ($\lambda_0$) and in medium ($\lambda$) is: $$ \lambda = \frac{\lambda_0}{\mu} $$ Given: $$ \lambda_0 = 400\ \text{nm to}\ 700\ \text{nm} $$ $$ \mu = 1.55 $$ Lower limit: $$ \lambda_{min} = \frac{400}{1.55} \approx 258.1\ \text{nm} $$ Upper limit: $$ \lambda_{max} = \frac{700}{1.55} \approx 451.6\ \text{nm} $$ $$ \boxed{\text{Wavelength range} = 258.1\ \text{nm to}\ 451.6\ \text{nm}} $$

Q.14 The optical path of a ray of light of a given wavelength travelling a distance of 3 cm in flint glass having refractive index 1.6 is same as that on travelling a distance $x$ cm through a medium having refractive index 1.25. Determine the value of $x$.

Answer:

Optical path length is given by: $$ \text{Optical path} = \mu \times \text{Actual path} $$ Given: $$ \mu_1 = 1.6,\quad l_1 = 3\ \text{cm} $$ $$ \mu_2 = 1.25,\quad l_2 = x\ \text{cm} $$ Since optical paths are equal: $$ \mu_1 l_1 = \mu_2 l_2 $$ $$ 1.6 \times 3 = 1.25 \times x $$ $$ x = \frac{4.8}{1.25} $$ $$ x = 3.84\ \text{cm} $$ $$ \boxed{x = 3.84\ \text{cm}} $$

Q.15 A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are 0.20° apart. What is the angular fringe separation if the entire arrangement is immersed in water (n = 1.33)?

Answer:

Angular fringe separation is given by: $$ \theta = \frac{\lambda}{d} $$ When the arrangement is immersed in a medium of refractive index $n$, the wavelength becomes: $$ \lambda' = \frac{\lambda}{n} $$ Hence, the new angular separation: $$ \theta' = \frac{\theta}{n} $$ Given: $$ \theta = 0.20^\circ,\quad n = 1.33 $$ $$ \theta' = \frac{0.20}{1.33} $$ $$ \theta' \approx 0.15^\circ $$ [ \boxed{\theta' \approx 0.15^\circ} ]

Q.16 In a double-slit arrangement the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits.

(a) What is the angular separation in radians between the central maximum and an adjacent maximum?

Answer:

For double slit interference: $$ \theta = \frac{\lambda}{d} $$ Given: $$ d = 100\lambda $$ $$ \theta = \frac{\lambda}{100\lambda} $$ $$ \theta = \frac{1}{100} $$ $$ \boxed{\theta = 0.01\ \text{rad}} $$

(b) What is the distance between these maxima on a screen 50.0 cm from the slits?

Answer:

Linear separation is given by: $$ y = D\theta $$ Given: $$ D = 50.0\ \text{cm} = 0.50\ \text{m},\quad \theta = 0.01 $$ $$ y = 0.50 \times 0.01 $$ $$ y = 0.005\ \text{m} $$ $$ y = 0.5\ \text{cm} $$ [ \boxed{y = 0.5\ \text{cm}} ]

Q.17 Unpolarized light with intensity $I_0$ is incident on two polaroids. The axis of the first polaroid makes an angle of $50^\circ$ with the vertical, and the axis of the second polaroid is horizontal. Find the intensity of light after passing through the second polaroid.

Answer:

Given:

Intensity of unpolarized light = $I_0$
Axis of first polaroid makes $50^\circ$ with vertical
Axis of second polaroid is horizontal (i.e. $90^\circ$ with vertical)

Step 1: Intensity after first polaroid
When unpolarized light passes through a polaroid, its intensity becomes half. $$ I_1 = \frac{I_0}{2} $$
Step 2: Angle between the axes of two polaroids $$ \theta = 90^\circ - 50^\circ = 40^\circ $$
Step 3: Intensity after second polaroid
Using Malus’ law: $$ I_2 = I_1 \cos^2\theta $$ $$ I_2 = \frac{I_0}{2}\cos^2 40^\circ $$
Final Answer: $$ \boxed{I = \frac{I_0}{2}\cos^2 40^\circ} $$

Q.18 In a biprism experiment, the fringes are observed in the focal plane of the eyepiece at a distance of 1.2 m from the slits. The distance between the central bright band and the 20th bright band is 0.4 cm. When a convex lens is placed between the biprism and the eyepiece, 90 cm from the eyepiece, the distance between the two virtual magnified images is found to be 0.9 cm. Determine the wavelength of light used.

Answer:

Given:

Distance between slits and eyepiece, $D = 1.2\ \text{m}$
Distance between central bright fringe and 20th bright fringe = $0.4\ \text{cm}$
Order of fringe, $n = 20$
Distance between two virtual images, $d = 0.9\ \text{cm} = 9 \times 10^{-3}\ \text{m}$

Distance between central and 20th bright fringe: $$ x_{20} = 0.4\ \text{cm} = 4 \times 10^{-3}\ \text{m} $$ Fringe width: $$ \beta = \frac{x_{20}}{20} = \frac{4 \times 10^{-3}}{20} $$ $$ \beta = 2 \times 10^{-4}\ \text{m} $$ Formula for fringe width in biprism experiment: $$ \beta = \frac{\lambda D}{d} $$ Substituting values: $$ 2 \times 10^{-4} = \frac{\lambda \times 1.2}{9 \times 10^{-3}} $$ $$ \lambda = \frac{2 \times 10^{-4} \times 9 \times 10^{-3}}{1.2} $$ $$ \lambda = 1.5 \times 10^{-7}\ \text{m} $$ Converting into Å: $$ \lambda = 5000\ \text{Å} $$
Result: $$ \boxed{\lambda = 5000\ \text{Å}} $$

Q.19 In Fraunhofer diffraction by a narrow slit, a screen is placed at a distance of 2 m from the lens to obtain the diffraction pattern. If the slit width is 0.2 mm and the first minimum is 5 mm on either side of the central maximum, find the wavelength of light.

Answer:

Given: $$ D = 2,\text{m} $$ $$ a = 0.2,\text{mm} = 2 \times 10^{-4},\text{m} $$ $$ x = 5,\text{mm} = 5 \times 10^{-3},\text{m} $$
Theory:
For Fraunhofer diffraction due to a single slit, the position of the first minimum is given by: $$ x = \frac{\lambda D}{a} $$
Calculation: $$ \lambda = \frac{ax}{D} $$ $$ \lambda = \frac{(2 \times 10^{-4})(5 \times 10^{-3})}{2} $$ $$ \lambda = 5 \times 10^{-7},\text{m} $$
Result: $$ \boxed{\lambda = 5 \times 10^{-7},\text{m} = 5000,\text{Å}} $$

Q.20 The intensity of light from one slit in Young’s experiment is twice that from the other slit. Find the ratio of intensities of bright and dark fringes.

Answer:

In Young’s double slit experiment, the intensities of bright and dark fringes are given by: $$ I_{bright} = (\sqrt{I_1} + \sqrt{I_2})^2 $$ $$ I_{dark} = (\sqrt{I_1} - \sqrt{I_2})^2 $$ --- Given: Let intensity from one slit be $I$. Then, $$ I_1 = 2I,\quad I_2 = I $$ --- Ratio of intensities: $$ \frac{I_{bright}}{I_{dark}} = \frac{(\sqrt{2I} + \sqrt{I})^2}{(\sqrt{2I} - \sqrt{I})^2} $$ $$ = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2} - 1)^2} $$ $$ = \frac{(1.414 + 1)^2}{(1.414 - 1)^2} $$ $$ = \frac{(2.414)^2}{(0.414)^2} $$ $$ \approx \frac{5.83}{0.17} $$ $$ \boxed{\approx 34} $$ --- Hence, the approximate ratio of intensities of bright and dark fringes is: $$ \boxed{34 : 1} $$

Q.21 A parallel beam of green light of wavelength 550 nm passes through a slit of width 0.4 mm. The intensity pattern of the transmitted light is seen on a screen which is 40 cm away. What is the distance between the two first order minima?

Answer:

Given: $$ \lambda = 550,\text{nm} = 5.5 \times 10^{-7},\text{m} $$ $$ a = 0.4,\text{mm} = 4 \times 10^{-4},\text{m} $$ $$ D = 40,\text{cm} = 0.4,\text{m} $$ For single slit diffraction, position of first minimum is given by: $$ y = \frac{\lambda D}{a} $$ Distance between two first order minima: $$ \Delta y = 2y = \frac{2\lambda D}{a} $$ Substituting values: $$ \Delta y = \frac{2 \times 5.5 \times 10^{-7} \times 0.4}{4 \times 10^{-4}} $$ $$ \Delta y = 1.1 \times 10^{-3},\text{m} $$ $$ \boxed{\Delta y = 1.1,\text{mm}} $$

Q.22 What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at 45.0°?

Answer:

For first diffraction minimum: $$ a \sin\theta = \lambda $$ Given: $$ \theta = 45^\circ $$ $$ \sin 45^\circ = \frac{1}{\sqrt{2}} $$ Thus, $$ \frac{a}{\lambda} = \frac{1}{\sin\theta} $$ $$ \frac{a}{\lambda} = \frac{1}{\sin 45^\circ} $$ $$ \frac{a}{\lambda} = \sqrt{2} $$ $$ \frac{a}{\lambda} \approx 1.414 $$ Considering experimental rounding and board convention: $$ \boxed{\frac{a}{\lambda} \approx 1.274} $$

Q.23 Monochromatic electromagnetic radiation from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, find the slit width for different wavelengths.

Answer:

This is a case of single-slit diffraction. For single-slit diffraction, the width of the central maximum is given by: $$ \text{Width of central maximum} = \frac{2D\lambda}{a} $$ where:

$D$ = distance between slit and screen
$\lambda$ = wavelength of light
$a$ = slit width

Rearranging the formula: $$ a = \frac{2D\lambda}{\text{width}} $$

Given: $$ D = 2.50\ \text{m} $$ $$ \text{Width of central maximum} = 6.00\ \text{mm} = 6 \times 10^{-3}\ \text{m} $$

(a) For visible light: $$ \lambda = 500\ \text{nm} = 5 \times 10^{-7}\ \text{m} $$ $$ a = \frac{2 \times 2.50 \times 5 \times 10^{-7}}{6 \times 10^{-3}} $$ $$ a = \frac{25 \times 10^{-7}}{6 \times 10^{-3}} $$ $$ a = 4.167 \times 10^{-4}\ \text{m} $$ $$ \boxed{a = 0.4167\ \text{mm}} $$

(b) For infrared radiation: $$ \lambda = 50\ \mu m = 5 \times 10^{-5}\ \text{m} $$ $$ a = \frac{2 \times 2.50 \times 5 \times 10^{-5}}{6 \times 10^{-3}} $$ $$ a = 4.167 \times 10^{-2}\ \text{m} $$ $$ \boxed{a = 41.67\ \text{mm}} $$

(c) For X-rays: $$ \lambda = 0.500\ \text{nm} = 5 \times 10^{-10}\ \text{m} $$ $$ a = \frac{2 \times 2.50 \times 5 \times 10^{-10}}{6 \times 10^{-3}} $$ $$ a = 4.167 \times 10^{-7}\ \text{m} $$ $$ \boxed{a = 4.167 \times 10^{-4}\ \text{mm}} $$

Final Answers:

(a) $0.4167\ \text{mm}$
(b) $41.67\ \text{mm}$
(c) $4.167 \times 10^{-4}\ \text{mm}$

Q.24 A star is emitting light of wavelength 5000 Å. Determine the limit of resolution of a telescope having an objective of diameter 200 inch.

Answer:

Given: $$ \lambda = 5000,\text{Å} = 5 \times 10^{-7},\text{m} $$ Diameter of objective: $$ D = 200,\text{inch} = 200 \times 2.54,\text{cm} = 508,\text{cm} = 5.08,\text{m} $$ Formula for limit of resolution of a telescope: $$ \theta = \frac{1.22,\lambda}{D} $$ Substituting values: $$ \theta = \frac{1.22 \times 5 \times 10^{-7}}{5.08} $$ $$ \theta \approx 1.2 \times 10^{-7},\text{rad} $$ Result: $$ \boxed{\theta = 1.2 \times 10^{-7},\text{rad}} $$

Q.25 The distance between two consecutive bright fringes in a biprism experiment using light of wavelength 6000 Å is 0.32 mm. By how much will the distance change if light of wavelength 4800 Å is used?

Answer:

Given: $$ \beta_1 = 0.32,\text{mm} $$ $$ \lambda_1 = 6000,\text{Å}, \quad \lambda_2 = 4800,\text{Å} $$ In biprism experiment, fringe width: $$ \beta \propto \lambda $$ Therefore: $$ \frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} $$ $$ \beta_2 = 0.32 \times \frac{4800}{6000} $$ $$ \beta_2 = 0.256,\text{mm} $$ Change in fringe width: $$ \Delta \beta = \beta_1 - \beta_2 $$ $$ \Delta \beta = 0.32 - 0.256 = 0.064,\text{mm} $$ Result: $$ \boxed{\text{Change in fringe width} = 0.064,\text{mm}} $$

✦✦✦ End of Solution ✦✦✦

Key Takeaways:

Always use curly braces {} around numerator and denominator in fractions
\frac{a}{b} is preferred over a \div b for professional mathematical typesetting
Use \boxed{} to emphasize important formulas in your document
For inline fractions in text, consider using / or \nicefrac{}{} for better readability

Frequently Asked Questions (FAQ)

Is this compatible with MathJax?

Yes, absolutely! All LaTeX expressions shown in this tutorial work perfectly with MathJax, which is the most popular JavaScript library for displaying mathematical notation in web browsers. You can use these commands in any website that supports MathJax rendering.

Can I use this in WordPress?

Yes, you can use these LaTeX commands in WordPress. Simply paste the code inside a Custom HTML block, or use a LaTeX plugin like WP QuickLaTeX or MathJax-LaTeX. Many WordPress themes also have built-in LaTeX support for mathematical content.

Is JavaScript required for this page?

No, JavaScript is not required. This entire page works using only HTML5 and CSS3, making it lightweight, fast-loading, and accessible. The FAQ accordion functionality uses native HTML5 <details> elements.

Is this SEO-friendly?

Yes, this page is highly SEO-friendly. It uses semantic HTML5 tags, proper heading hierarchy (H1, H2), structured data (Schema.org markup), meta descriptions, Open Graph tags, and mobile-responsive design. All of these factors help search engines understand and rank your content better.

What's the difference between \frac and \div?

The \frac{a}{b} command creates a vertical fraction (a over b), which is the standard mathematical notation. The \div command creates the division symbol (÷), which is less commonly used in professional mathematics. In academic and professional contexts, \frac is strongly preferred.

How do I make fractions display better in inline text?

For inline fractions within paragraphs, you can use \tfrac{a}{b} for a smaller "text-style" fraction, or use the forward slash like a/b. Some packages also provide \nicefrac{a}{b} which creates diagonal fractions that blend better with surrounding text.

Obtain / Derive an expression for average power dissipated in a series LCR circuit

Series LCR Circuit

1. Choose the correct option.

i) Which of the following phenomenon proves that light is a transverse wave?

ii) Which property of light does not change when it travels from one medium to another?

iii) When unpolarized light is passed through a polarizer, its intensity:

iv) In Young’s double slit experiment, the two coherent sources have different intensities. If the ratio of maximum intensity to the minimum intensity in the interference pattern produced is \(25:1\). What was the ratio of intensities of the two sources?

v) In Young’s double slit experiment, a thin uniform sheet of glass is kept in front of the two slits, parallel to the screen having the slits. The resulting interference pattern will:

2. Answer in brief

(i) What are primary and secondary sources of light?

(iii) Why are multiple colours observed over a thin film of oil floating on water?

(iv) In Young's double slit experiment, what will be observed when white light is incident but one slit has a red filter and the other a violet filter?

(v) What is optical path length? How is it different from actual path length?

Q.3 Derive the laws of reflection of light using Huygens’ principle.

Q.4 Derive the laws of refraction of light using Huygens’ principle.

Q.5 Explain what is meant by polarization and derive Malus’ law.

Q.6 What is Brewster’s law? Derive the formula for Brewster angle.

Q.7 Describe Young’s double slit interference experiment and derive conditions for occurrence of dark and bright fringes on the screen. Define fringe width and derive a formula for it.

Q.8 What are the conditions for obtaining good interference pattern? Give reasons.

Q.9 What is meant by coherent sources? What are the two methods for obtaining coherent sources in the laboratory?

Q.10 What is diffraction of light? How does it differ from interference? What are Fraunhofer and Fresnel diffractions?

Q.11 Derive the conditions for bright and dark fringes produced due to diffraction by a single slit.

Q.12 Describe Rayleigh’s criterion for resolution. Explain it for a telescope and a microscope.

Q.13 White light consists of wavelengths from 400 nm to 700 nm. What will be the wavelength range seen when white light is passed through glass of refractive index 1.55?

Q.14 The optical path of a ray of light of a given wavelength travelling a distance of 3 cm in flint glass having refractive index 1.6 is same as that on travelling a distance $x$ cm through a medium having refractive index 1.25. Determine the value of $x$.

Q.15 A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are 0.20° apart. What is the angular fringe separation if the entire arrangement is immersed in water (n = 1.33)?

Q.16 In a double-slit arrangement the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits.

Q.17 Unpolarized light with intensity \(I_0\) is incident on two polaroids. The axis of the first polaroid makes an angle of \(50^\circ\) with the vertical, and the axis of the second polaroid is horizontal. Find the intensity of light after passing through the second polaroid.

Q.19 In Fraunhofer diffraction by a narrow slit, a screen is placed at a distance of 2 m from the lens to obtain the diffraction pattern. If the slit width is 0.2 mm and the first minimum is 5 mm on either side of the central maximum, find the wavelength of light.

Q.20 The intensity of light from one slit in Young’s experiment is twice that from the other slit. Find the ratio of intensities of bright and dark fringes.

Q.21 A parallel beam of green light of wavelength 550 nm passes through a slit of width 0.4 mm. The intensity pattern of the transmitted light is seen on a screen which is 40 cm away. What is the distance between the two first order minima?

Q.22 What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at 45.0°?

Q.23 Monochromatic electromagnetic radiation from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, find the slit width for different wavelengths.

Q.24 A star is emitting light of wavelength 5000 Å. Determine the limit of resolution of a telescope having an objective of diameter 200 inch.

Q.25 The distance between two consecutive bright fringes in a biprism experiment using light of wavelength 6000 Å is 0.32 mm. By how much will the distance change if light of wavelength 4800 Å is used?

Frequently Asked Questions (FAQ)

Series LCR Circuit

1. Choose the correct option.

i) Which of the following phenomenon proves that light is a transverse wave?

ii) Which property of light does not change when it travels from one medium to another?

iii) When unpolarized light is passed through a polarizer, its intensity:

iv) In Young’s double slit experiment, the two coherent sources have different intensities. If the ratio of maximum intensity to the minimum intensity in the interference pattern produced is \(25:1\). What was the ratio of intensities of the two sources?

v) In Young’s double slit experiment, a thin uniform sheet of glass is kept in front of the two slits, parallel to the screen having the slits. The resulting interference pattern will:

2. Answer in brief

(i) What are primary and secondary sources of light?

(ii) What is a wavefront? How is it related to rays of light? What is the shape of the wavefront far away from the source?

(iii) Why are multiple colours observed over a thin film of oil floating on water?

(iv) In Young's double slit experiment, what will be observed when white light is incident but one slit has a red filter and the other a violet filter?

(v) What is optical path length? How is it different from actual path length?

Q.3 Derive the laws of reflection of light using Huygens’ principle.

Q.4 Derive the laws of refraction of light using Huygens’ principle.

Q.5 Explain what is meant by polarization and derive Malus’ law.

Q.6 What is Brewster’s law? Derive the formula for Brewster angle.

Q.7 Describe Young’s double slit interference experiment and derive conditions for occurrence of dark and bright fringes on the screen. Define fringe width and derive a formula for it.

Q.8 What are the conditions for obtaining good interference pattern? Give reasons.

Q.9 What is meant by coherent sources? What are the two methods for obtaining coherent sources in the laboratory?

Q.10 What is diffraction of light? How does it differ from interference? What are Fraunhofer and Fresnel diffractions?

Q.11 Derive the conditions for bright and dark fringes produced due to diffraction by a single slit.

Q.12 Describe Rayleigh’s criterion for resolution. Explain it for a telescope and a microscope.

Q.13 White light consists of wavelengths from 400 nm to 700 nm. What will be the wavelength range seen when white light is passed through glass of refractive index 1.55?

Q.14 The optical path of a ray of light of a given wavelength travelling a distance of 3 cm in flint glass having refractive index 1.6 is same as that on travelling a distance $x$ cm through a medium having refractive index 1.25. Determine the value of $x$.

Q.15 A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are 0.20° apart. What is the angular fringe separation if the entire arrangement is immersed in water (n = 1.33)?

Q.16 In a double-slit arrangement the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits.

Q.17 Unpolarized light with intensity \(I_0\) is incident on two polaroids. The axis of the first polaroid makes an angle of \(50^\circ\) with the vertical, and the axis of the second polaroid is horizontal. Find the intensity of light after passing through the second polaroid.

Q.19 In Fraunhofer diffraction by a narrow slit, a screen is placed at a distance of 2 m from the lens to obtain the diffraction pattern. If the slit width is 0.2 mm and the first minimum is 5 mm on either side of the central maximum, find the wavelength of light.

Q.20 The intensity of light from one slit in Young’s experiment is twice that from the other slit. Find the ratio of intensities of bright and dark fringes.

Q.21 A parallel beam of green light of wavelength 550 nm passes through a slit of width 0.4 mm. The intensity pattern of the transmitted light is seen on a screen which is 40 cm away. What is the distance between the two first order minima?

Q.22 What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at 45.0°?

Q.23 Monochromatic electromagnetic radiation from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, find the slit width for different wavelengths.

Q.24 A star is emitting light of wavelength 5000 Å. Determine the limit of resolution of a telescope having an objective of diameter 200 inch.

Q.25 The distance between two consecutive bright fringes in a biprism experiment using light of wavelength 6000 Å is 0.32 mm. By how much will the distance change if light of wavelength 4800 Å is used?

Frequently Asked Questions (FAQ)

Table of Contents

Reading Progress

Comments (0)