WBCHSE Class 12 Physics Light Polarisation Questions and Answers

Optics

Diffraction And Polarisation Of Light Long Question And Answers

Question 1. How will the diffraction pattern in a single slit be affected if

1. The width of the slit is increased and
2. The wavelength of the incident light is increased?

Answer:

The angular width of a diffraction pattern is given by θ= \(\frac{\lambda}{a}\), where X = wavelength of the incident light and a of the slit

If the width of the slit a is increased, θ will decrease. So the diffraction bands will come closer to each other. On increasing the slit width, at a certain value of a, the diffraction pattern will be no more distinct to observe.

If the wavelength of the incident light A is increased, the angular width of the pattern will increase. Hence the diffraction bands will be wider

Question 2. What will be the effect on the diffraction pattern in a single slit if red light is used instead of violet light?
Answer:

We know that the angular width of diffraction in a single slit is given by θ= \(\frac{\lambda}{a}\)

Now, the wavelength of red light (λ_r) is greater than that of violet (λ_ν) i.e λ_r>λ_ν

So, θ_r>θ_ν

Therefore diffraction bands in red light will be wider

Question 3. Radio waves diffract strongly around big buildings but light waves do not. Why? 
Answer:

The wavelength range of 4 ×10^-7 m to 7 ×10^-7 m while that Qf radio waves is from 0.1 m to 10 4 i.e. the wavelength of radio waves is much greater than that of light waves. We know that the bending of waves increases with the increase of its wavelength. So radio waves are easily diffracted while the wavelength of light waves is too small to be diffracted around big buildings

Question 4. If a single slit is illuminated by white light, what will be the nature of color of the diffraction pattern?
Answer:

If a single slit is illuminated by white light, the central maximum of the diffraction pattern is produced due to the reinforcement of the diffracted waves having the same phase but different wavelengths. So the central maximum will be of white color. But in the case of secondary maxima on either side of the central maximum, the angle of diffraction θ ∝\(\) is taken very small.

Since λ_r >λ_ν,   So, θ_r >θ_ν

Hence the secondary maxima will be coloured. The inner portion of each maximum will be of violet color and the portion will be of red color

WBBSE Class 12 Polarisation Q&A

Question 5. A tourmaline crystal is placed in the path of a polarised beam of light. If it is rotated through one complete rotation, what change will be observed in the intensity of the transmitted light?
Answer:

The optic axis of the tourmaline crystal is placed along the direction of polarisation of the polarised light. The light will be transmitted through the crystal and its intensity will remain unchanged. Now taking the light ray as the axis of rotation of the crystal, it is rotated through one complete rotation.

The observations are as follows:

• The intensity will gradually diminish and become zero when the crystal is rotated through 90°
• Next intensity begins to increase and becomes equal to its initial value when the crystal is rotated through 180°
• Again intensity begins to decrease and becomes zero when the crystal is rotated through 270° and
• finally, intensity begins to increase and becomes equal to its initial value when the crystal is rotated through 360°

Question 6. How will you identify experimentally whether a given beam oflight is plane polarised or unpolarised?
Answer:

A tourmaline crystal is placed in the path of. the given beam oflight and it is rotated.If the intensity of the transmitted light through the crystal remains unchanged, the beam is unpolarised. Intensity varies periodically and becomes zero twice in each rotation, then the beam is plane-polarised

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Question 7. Show that the width of the central maximum is twice that of a secondary maximum and if the width of slit is increased, the width of the diffraction fringes gets diminished.
Answer:

Distance of nth minimum from central maximum,

⇒ \(x_n=\frac{n D \lambda}{a}\)

[Where D = distance of slit from the screen, λ = wavelength of the light, a = width of the slit]

In a diffraction pattern, secondary maxima and minima alternatively and so the width of a secondary maximum

⇒ \(\beta=x_n-x_{n-1}=\frac{n D \lambda}{a}-\frac{(n-1) D \lambda}{a}=\frac{D \lambda}{a}\)

Now, the angular width of the central maximum

⇒  \(2 \theta=\frac{2 \lambda}{a}\)

Therefore, the linear width central maximum

⇒ \(\beta_0=D \cdot 2 \theta=\frac{2 D \lambda}{a}\) = 2β

i.e., the width of the central maximum is twice that of any secondary maximum

Again, both the width of the secondary maximum and central

⇒  \(\text { maximum } \propto \frac{1}{\text { width of }{slit}(a)}\)

So, if the width of slit is increased, the width of diffraction fringes gets diminished

Short Answer Questions on Polarisation of Light

Question 8. Results of two experiments on single-slit diffraction

How can the ratio of widths of two slits used in the experiments be evulated from the given results?
Answer:

Let us assume the width of the two slits, used in the experiments are d and d’

⇒ \(\theta=\frac{\lambda}{d}\)

And \(q \theta=\frac{p \lambda}{d^{\prime}}\)

⇒ \(\frac{d}{d^{\prime}}=\frac{q}{p}\)

So, the ratio of widths of two slits = d: d’ – q : P

Question 9. If the diameter of the objective of a telescope is doubled, how will its resolving power change?
Answer:

Resolving power of telescope, R = \(\frac{a}{1.22 \lambda}\)

The revolving power of a telescope changes in direct proportion to the diameter of its objective. So,if the diameter of the objective of a telescope is increased two folds, its resolving power will also be increased twofold

Question 10. If a ray of light Is incident on a reflecting medium at the polarising angle, then prove that the reflected and the refracted rays are at 90° to each other
Answer:

Let a ray of light AO be incident on XY at the polarising angle p. The reflected ray OB which is plane polarised, also makes an angle p with the normal.

Let the angle of refraction be r

According to Brewster’s law, tan p = p [where p = refractive index of the medium]

According to the law of refraction

mu = \(\frac{\sin \angle A O N}{\sin \angle N^{\prime} O C}=\frac{\sin p}{\sin r}\)

tan p = \(\frac{\sin p}{\sin r}\)

Or, \(\frac{\sin p}{\cos p}=\frac{\sin p}{\sin r}\)

or, sin r= cosp = sin (90°- p) or, r = 90°-polarising

r+p = 90°

So, the reflected ray OB and the refracted ray OC are at 90° to each other

Question 11. The critical angle between a and air is θ_c. A ray of light In air enters the transparent. medium at an angle of Incidence equal to the polarising angle. Deduce a relation between the angle of refraction r and critical angle θ_c. given transparent medium
Answer:

Let the refractive index of the transparent medium polarising angle = ip

So, \(\mu=\frac{\sin i_p}{\sin r}\)

Again, i_p+i_r= 90°

sin i_p= sin (90°-r) = cost

⇒ \(\mu=\frac{\cos r}{\sin r}\)

According to SneU’s law, for critical angle 6

⇒ \(\mu=\frac{1}{\sin \theta_c}\)

⇒ \(\frac{\cos r}{\sin r}=\frac{1}{\sin \theta_c}\)

Or, \(\frac{\sin r}{\cos r}\)

Or, tan r = sin θ_c

Or, tan^-1 (sin θ_c)

This is the required relation.

Common Questions on Polarising Filters

Question 12. Deduce a relation between the polarising angle and the critical angle.
Answer:

According to Brewster’s law

μ = tan i_p = Where i_p = polarising angle]

Again, according to Snell’s law

⇒ \(\mu=\frac{1}{\sin \theta_c}\)

Where = θ_c critical angle

⇒ \(\tan i_p=\frac{1}{\sin \theta_{\mathrm{c}}}\)

Or, tan i_p  = cosec θ_c

or, i_p = tan^-1 (cosec θ_c )

This is the required relation.

Question 13. The optic axes of two polaroids are inclined at an angle of 45° with each other. Unpolarised light of intensity IQ being incident on the first polaroid emerges from the second polaroid. Find the intensity of the emergent light.
Answer:

In the first polaroid, the component of incident unpolarised light having polarisation parallel t6 the optic axis is transmitted through the polaroid, and the perpendicular component is absorbed. So the intensity of light incident on the second polaroid \(\frac{J_0}{2}\)

Again, the intensity of light ∝ (amplitude of light)²

If the amplitude of light incident on the second polaroid is Eo , then the amplitude of light having polarisation along its optic axis

⇒ \(E_0 \cos 45^{\circ}=\frac{E_0}{\sqrt{2}}\)

The angle between the optics axes = 45°

So the intensity of light transmitted through an incident on it

Hence the intensity of the emergent light = \(\frac{1}{2} \cdot \frac{I_0}{2}=\frac{I_0}{4}\)

Question 14. An unpolarised light is incident at the angle of polarization on a reflector. Determine the angle between the reflected and the transmitted rays

μ = tan i_p [i_p =a angle of polarisation]

Again  μ = \(\frac{\sin i_p}{\sin r}\)

Or, tan i_p = \(\frac{\sin i_p}{\sin r}\)

Or, in r = cos i_p = sin (90°- i_p)

Or, r = 90° – i_p

Or, i_p+ r= 90°

From the figure, the angle between the reflected and the transmitted rays

= 180° – (i_p + r)  100° -90° =90°

Practice Questions on Brewster’s Law

Question 15.

1. How does the angular width of the central maxima in a single-slit Fraunhofer diffraction experiment change when the distance between the silt and screen is doubled?
2.  In the Fraunhofer diffraction experiment, the first minima of red light = 660 ntn) is formed on the first maxima of another light of wavelength A’. Find the value of A‘

Answer:

The angular width of the central maxima will not change because it does not depend on the distance between the slit and the screen.

The angle of diffraction for the first minima

θ = \(=\frac{\lambda}{a}\)

The angle of diffraction for the first maxima

⇒ \(\theta^{\prime}=(2 n+1) \frac{\lambda^{\prime}}{2 a}=(2 \times 1+1) \frac{\lambda^{\prime}}{2 a}=\frac{3 \lambda^{\prime}}{2 a}\)

Given, θ  = θ ‘

⇒ \(\frac{\lambda}{a}=\frac{3 \lambda^{\prime}}{2 a} \quad \text { or, } \lambda^{\prime}=\frac{2}{3} \lambda\)

Question 16. Two polaroids A and B are kept in a crossed position. How should a third polaroid C be placed between them so that the intensity of polarised light transmitted by polaroid B reduces to 1/8 th of the intensity of unpolarised light incident on A
Answer:

Intensity of incident = IQ

Intensity of light after passing through polaroid

A = \(\frac{I_0}{2}\)

Intensity of light after passing through polaroid

C = \(\frac{I_0}{2} \cos ^2 \theta\)

Where θ is the angle between the pass axis of the A and C polaroid.

∴ Intensity of light after passing through B,

I = \(\frac{I_0}{2} \cos ^2 \theta \cos ^2\left(90^{\circ}-\theta\right)=\frac{I_0}{8} \sin ^2 2 \theta\)

If \(I=\frac{I_0}{8}\), then sin² 2θ = i Or, θ   = 45°

So thepolaroid C shouldbeplacedmaking an angle  45° with the pass axis ofpolaroid A

Question 17. A parallel beam of light of 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away.It is observed that the first minimum is at a distance of 2.5 mm from the center of the screen. Calculate the width of the site
Answer:

Wavelength of incident light, A = 500 nm = 500 x 10-9 m Distance between slit and screen, D = 1 m For 1st minima, angular width on either side of central maxima = \(\)

⇒ \(\frac{\lambda}{a}\)

a = \(\frac{\lambda}{a} \cdot D=2.5 \times 10^{-3} \mathrm{~m}\)

a = \(\frac{\lambda \cdot D}{2.5 \times 10^{-3}}=\frac{500 \times 10^{-9} \times 1}{2.5 \times 10^{-3}}\)

= 2 × 10^-4 m

= 0.2 mm

Important Definitions in Light Polarisation

Question 18. Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2  × 10^-6 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases
Answer:

Angular position of first maxima, x = \(\frac{\lambda D}{a}\)

Where, D = 1.5 m, a = 2 × 10^-6 m

For λ = 590 nm,

x₁ = \(\frac{596 \times 10^{-9} \times 1.5}{2 \times 10^{-6}}\)

= 0.4425m

= 44.25cm

For λ = 596 nm

x₂ =\(\frac{596 \times 10^{-9} \times 1.5}{2 \times 10^{-6}}\)

= 0.447m

= 44.7 cm

∴ Separation = x₂ -x₁ = 0.45 cm

Question 19.

1. Why does unpolarized light from a source show a variation in intensity when viewed through a period that is rotated? Show with the help of a diagram, how unpolarised light from the sun gets linearly polarised by scattering.

2. Three identical polaroid sheets P₁, P₂, and P₃ are oriented so that the pass axis of P₂ and P₃ are inclined at § angles of 60° and respectively with the pass axis of O 90° P₁. A monochromatic source S of unpolarized light of intensity IQ is kept in front of the polaroid sheet P₂. Determine the intensities of light as observed by the observer at O. When polaroid P₃ is rotated with respect to P₂ at angles θ = 30° and 60°

Answer:

1. Polaroid films are produced by spreading ultramicroscopic crystals of her pathic on a thin sheet of nitrocellulose. The crystals are placed on the film by a special device in such a way that the optic axes of all the crystals are parallel. These crystals are

Highly dichroic and absorbs one of the doubt-refracted beams completely. The other refracted beam is transmitted from the crystals.

2. The ray of light passing through polaroid Px will have intensity reduced by half

⇒ \(I_1=\frac{I_0}{2}\)

Now, the polaroid P₂ is oriented at an angle 60° respect to with

⇒ \(I_2=I_1 \cos ^2 60=\frac{I_0}{2} \times \frac{1}{4}=\frac{I_0}{8}\)

Now, the polaroid P₃ is originally oriented at an angle

90° – 60°  = 30°.

Hence, when P₃ is rotated by 30°, the angle between

P₂ and P₃ is 60°.

Therefore, the intensity

⇒ \(I_3=I_2 \cos ^2 60\)

⇒ \(\frac{I_0}{8} \cos ^2 60\)

⇒ \(\frac{I_0}{8} \times \frac{1}{4}=\frac{I_0}{32}\)

Similarly, when P3 is rotated by 60°, the angle between

P₂ and P₃ is 90°.

Therefore, the intensity is

⇒ \(I_3^{\prime}=I_2 \cos ^2 90\)

⇒ \(\frac{I_0}{8} \times 0\)

= 0

Examples of Applications of Light Polarisation

Question 20. A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen. Estimate the number of fringes obtained in Young’s double slit experiment with fringe width 0.5 mm .which can be accommodated within the region of total angular spread of the central maximum due to single slit
Answer: 

λ = 500 nm = 5 ×10^-7 m,

a = 0.2 mm = 2 ×10^-4 m

Angular width of central maxima

⇒ \(\frac{2 \lambda}{a}=\frac{2 \times 5 \times 10^{-7}}{2 \times 10^{-4}}\)

⇒ \(5 \times 10^{-3}\)

The fringe width in Young’s double slit experiment,

β = 0.5 mm = 5 ×10^-4  m

The number of fringes obtained

N = \(\frac{\theta_0 D}{\beta}=\frac{5 \times 10^{-3} \times 1}{5 \times 10^{-4}}\)

= 10 Assuming D = 1m

Question 21. Unpolarised light is passed through a polaroid P₁. When this polarised beam passes through another polaroid P₂ and if the pass axis of P₂ makes an angle θ with the pass axis of P₁, then write the expression for the polarised beam passing through P₂. Draw a plot showing the variation of intensity when θ varies from 0 to 2π.

Let the intensity unpolarised light incident on Py be IQ.

Then the intensity of the light

In transmitted by P₁ = \(\frac{I_0}{2}\)

Applying Malus’ law, the intensity of light transmitted

By P₂ = \(\frac{I_0}{2} \cos ^2 \theta\)

Real-Life Scenarios in Polarisation Experiments

Question 22. How is linearly polarised light obtained by the process of scattering of light? Find the Brewster angle for the air-glass interface, when the refractive index of glass = 1.5.
Answer:

Sunlight scattered by the sky is polarised. This is because the molecules of the Earth’s atmosphere acquire motion in two mutually perpendicular directions when sunlight falls on them, but an observer on Earth receives light only from those molecules that move in the transverse direction

The Brewster angle for the air-glass interface

= tan^-1μ= tan^-1 1.5 = 56.30°

Question 23.

1. In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band? Explain
Answer: 

2. When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot Is seen at the center of the obstacle. Explain why

Answer: 

The width of the central diffraction band = \(\frac{2 D \lambda}{a}\)

1. When the width of the slit is doubled, the size of the central diffraction band is halved, and its intensity increases.

2. Waves diffracted from the edge of the circular obstacle produce constructive interference at the tire center and form a bright spot

Question 24. How does the resolving power of a microscope depend on

1. The wavelength of the light used and
2. The medium used between the object and the objective lens?

Resolving power, R = \(\frac{2 \mu \sin \theta}{\lambda}\)

The resolving power of a microscope varies inversely with the wavelength of the light used.

The resolving power of a microscope is directly proportional to the refractive index of the medium used between the object and the objective lens.

Conceptual Questions on Types of Polarisation

Question 25. Unpolarised light incident from air on a plane surface of a material of refractive index μ. At a particular angle of incidence i, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?

i = \(\sin ^{-1}\left(\frac{1}{\mu}\right)\)

Reflected light is polarised with its electric vector perpendicular to the plane of incidence

i = \(\tan ^{-1}\left(\frac{1}{\mu}\right)\)

The reflected and refracted rays are perpendicular to each other when light incident at at Brewster’s angle

Hence the reflected light is polarised with its electric vector perpendicular to the plane of incidence.