WBCHSE Class 12 Physics Light Polarisation Short Answer Questions

Optics

Diffraction And Polarisation Of Light Short Question And Answers

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

The width ofthe central maxima = \(\frac{2 \lambda D}{d}\) . When d is doubled the width reduces to half.

Question 2. When a small circular obstacle is placed in front of a white wall in the path of light rays coming from a distant source, bright spots are observed at the center of the shadow. Explain.
Answer:

Diffraction of light waves takes place at the edges of the circular obstacle. Constructive interference of these diffracted rays gives rise to the bright spot at the center of the shadow on the wall.

Question 3. In a 10 m high room a partition of height 7 m separates two students on either side. Both light and sound waves can deviate from their path if they experience any obstruction. Then why is it that the two students can converse with each other even if one cannot see the other?
Answer:

For diffraction to occur, the size of the obstacle should be comparable to the wavelength. The wavelength of sound waves(≈ 0.33 m) is much larger than the wavelength of light.

Waves ((≈ 10^-7 m) sound waves can be diffracted through the edges of much larger obstacles like walls or partitions. On the other hand, the wavelength of light is much smaller than the height of the partition making diffraction impossible. So the two boys can hear but cannot see each other.

Question 4. Ray optics is based on the assumption that light travels in a straight line. Diffractions observed when light propagates through small apertures/slits or around small objects disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?
Answer:

The apertures of optical instruments are much larger than the wavelength of the light passing through them. So there is no possibility of any diffraction taking place and there is no anomaly in the wave optics and ray optics in all practical purposes.

Question 5. A parallel beam of light of wavelength 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. Find the width of the slit. 
Answer: 

For the first darkband, sin = \(\frac{\lambda}{d}\)

But, = \(\frac{2.5 \times 10^{-3}}{1}\)

= \(\frac{\lambda}{\sin \theta_1}=\frac{500 \times 10^{-9}}{2.5 \times 10^{-3}}\)

= 0.2 mm

Question 6. What is understood by the diffraction of light? In the n single slit experiment, if the width of the slit increases, what will be the change of the angular width of the central maxima? State Brewster slaw. 

The angular width (2θ) offline central maximum decreases in the same ratio at which the width of the slit increases.

Question 7. The resolving power of a microscope at 6000 A° is 10⁴. What resolving power at 4000 A°
Answer:

The resolving

⇒ \(R \propto \frac{1}{\lambda}\)

So, \(\frac{R_1}{R_2}=\frac{\lambda_2}{\lambda_1}\)

Or, \(R_2=R_1 \frac{\lambda_1}{\lambda_2}\)

⇒ \(R_2=10^4 \times \frac{6000}{4000}=1.5 \times 10^4\)

Question 8. The critical angle of a transparent crystal for green light is 30°. Find the angle of polarization of that crystal.

If the angle of polarisation is i_p then

tan i_p = mu \(\) = 2

Or, i_p= tan^-1(2) = 63.435°

Question 9. Why are polaroids used In sunglasses?
Answer:

Unpolarised light is polarised by a polaroid. The polarising axis is kept horizontal in a sunglass so that the light is comforting for the eye

Question 10. How does the angular separation between fringes in a single slit diffraction experiment change when the distance of separation between the slit and screen is doubled?

The angular separation between fringes in a single slit diffraction pattern does not change with the distance between the slit and screen

Question 11. 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?
Answer: 

The line width of the central diffraction fringe \(\propto \frac{1}{\text { slit width }}\)  The slit width is made double then the width of the central fringe becomes half of the initial value and intensity will increase

Question 12. In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
Answer:

The diffraction pattern of each slit modulates the intensity of interference fringes in a double-slit arrangement.

Question  13. 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 shadow of the obstacle. Explain why.
Answer:

When a tiny circular obstacle is placed in the path of light from a distant source, waves diffracted from the edge of the circular obstacle interfere constructively at the center of the shadow, producing a right spot.

Question 14. The light coming from a portion of the sky is sunlight that has changed its direction due to scattering by molecules hr the earth’s atmosphere. This scattered light is polarised. Therefore, It shows a variation in intensity when viewed through a polaroid on rotation
Answer: 

The light coming from a portion of the sky is sunlight that has changed its direction due to scattering by molecules hr the earth’s atmosphere. This scattered light is polarised. Therefore, It shows a variation in intensity when viewed through a polaroid on rotation.

Question 15. When are two objects just resolved? Explain. How can the resolving power of a compound microscope be increased? Use relevant formulas to support your answer.
Answer:

Two objects are just resolved by an optical system when the central maximum of the diffraction pattern due to one falls on the first minimum of the diffraction pattern of the other

Question 16. A narrow beam of unpolarised light of intensity I₀ is incident on a polaroid P₁. The light transmitted by it is then incident on a second polaroid P₂ with its pass axis making an angle of 60° relative to the pass axis of P₁. Find the intensity of the light transmitted by P₂
Answer:

Intensity of light transmitted by P1= \(\frac{I_0}{2}\)

Applying Malus’ law, the intensity of light transmitted by

P₂= \(\frac{I_0}{2} \cos ^2 60^{\circ}=\frac{I_0}{8}\)