Young's experiment is performed in air and then performed in water, the fringe width 

1. Will remain same

2. Will decrease

3. Will increase

4. Will be infinite

In the Young's double slit experiment, if the phase difference between the two waves interfering at a point is $\mathrm{\varphi }$, the intensity at that point can be expressed by the expression

1. $\mathrm{l}=\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2 {\cos }^2\mathrm{\varphi }}$

2. $\mathrm{I}=\frac{\mathrm{A}}{\mathrm{B}}\mathrm{cos\varphi }$

3. $\mathrm{I}=\mathrm{A}+\mathrm{B} \cos \frac{\mathrm{\varphi }}{2}$

4. $\mathrm{I}=\mathrm{A}+\mathrm{B} \mathrm{cos\varphi }$

Where A and B depend upon the amplitude of the two waves.

A slit of width a is illuminated by white light. For red light $\left(\mathrm{\lambda }=6500\stackrel{°}{\mathrm{A}}\right)$, the first minima is obtained at $\mathrm{\theta }=30°$. Then the value of a will be 

1. $3250 \stackrel{°}{\mathrm{A}}$

2. $6.5\times {10}^{-4} \mathrm{mm}$

3. 1.24 microns

4. $2.6\times {10}^{-4} \mathrm{cm}$

Two waves originating from source S₁ and S₂ having zero phase difference and common wavelength $\lambda$ will show completely destructive interference at a point P if $\left({\mathrm{S}}_1\mathrm{P}-{\mathrm{S}}_2\mathrm{P}\right)$ is

1. 5$\lambda$

2. 3$\lambda$ / 4

3. 2$\lambda$

4. 11$\lambda$ / 2

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