The Brewster angle for the glass-air interface is 54.74^o. lf a ray of light going from air to glass strikes at an angle of incidence 45^o, then the angle of refraction is

(Hint: tan 54.74^o = $\sqrt{2}$)

1. 60^o

2. 30^o

3. 25^o

4. 54.74^o

The path difference between two interfering waves at a point on the screen is $\lambda /6.$ The ratio of intensity at this point and that at the central bright fringe will be (Assume that intensity due to each slit in same)

1. 0.853

2. 8.53

3. 0.75

4. 7.5

The equation of two light waves are $y_1=6 cos\omega t, {\mathrm{y}}_2=8\cos \left(\omega t+\varphi \right).$ The ratio of maximum to minimum intensities produced by the superposition of these waves will be -

1. 49: 1

2. 1: 49

3. 1: 7

4. 7: 1

In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $\mathrm{\lambda }$ is K, ($\mathrm{\lambda }$ being the wavelength of light used). The intensity at a point where the path difference is $\frac{\mathrm{\lambda }}{4}$, will be

1. K/2

2. Zero

3. K

4. K/4

In Young's double slit experiment intensity at a point is (1/4) of the maximum intensity. Angular position of this point is 

1. ${\sin }^{-1}\left(\lambda /d\right)$

2. ${\sin }^{-1}\left(\lambda /2d\right)$

3. ${\sin }^{-1}\left(\lambda /3d\right)$

4. ${\sin }^{-1}\left(\lambda /4d\right)$

A light has amplitude A and angle between analyser and polariser is 60^o. Light is reflected by analyser has amplitude

1. A$\sqrt{2}$

2. A/$\sqrt{2}$

3. $\sqrt{3}$A/2

4. A/2

The intensity ratio of two waves is 9: 1 . These waves produce the event of interference. The ratio of maximum to minimum intensity will be-

1. 1: 9

2. 9: 1

3. 1: 4

4. 4: 1

If an interference pattern has maximum and minimum intensities in 36: 1 ratio then what will be the ratio of amplitudes

1. 5: 7

2. 7: 4

3. 4: 7

4. 7: 5

Light of wavelength 589.3 nm is incident normally on the slit of width 0.1 mm. What will be the agular width of the central diffreaction maximum at a distance of 1 m from the slit

1. $0.68°$

2. $1.02°$

3. $0.34°$

4. None of these

In a single-slit diffraction experiment, the width of the slit is reduced by half. Which of the following needs to be done if the width of the central maxima has to remain the same

1. Reduce the distance between the slit and screen by half

2. Reduce the distance between the slit and the screen to ${\left(\frac{1}{4}\right)}^{th}$ the original separation

3. Double the distance between the slit and the screen

4. No need to do anything, as the width of the central maxima does not depend on the slit width