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**Q.1. Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2m is: (2020)A: 7.32 × 10^{–7} rad **

C: 3.66 × 10

D: 1.83 × 10

Ans:

A: 45° < i

B: i

C: 0° < i

D: 30° < i

Ans:

m = tan (i

i

(μ > 1)

i

45 < i

A: four times

C: double

D: half

Fringewidth in YDSE is given by

A: 0.266°

B: 0.15°

C: 0.05°

D: 0.1°

Ans:

In air angular fringe width

Angular fringe width in water

A: Reflected light is polarised with its electric vector parallel to the plane of incidence

**Q.6. ****In Young's double slit experiment the separation between the slits is 2 mm, the wavelength λ of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same λ and D) the separation between the slits needs to be changed to :- (2018)****A: 1.8 mmB: 1.9 mmC: 2.1 mmD: 1.7 mm**

A: 1.59

B: 1.69

C: 1.78

D: 1.25

Ans:

Solution:

Fringe Width in a medium -

- where in

Since wavelength in a medium become

Fringe width in air

Fringe width in any other medium

According to question

A:

B:

C:

D:

**Ans: **B

Solution:

**Q.9. ****In a diffraction pattern due to a single slit of width 'a' the first minimum is observed at an angle 30º when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of : (2016)A: sin-1 (3/4)B: sin-1 (1/4)C: sin-1 (2/3)D: sin-1 (3/2)Ans: **A

Solution:

Given that, first minimum is observed at an angle of 30º in a diffraction pattern due to a single slit of width a.

i.e., n = 1, θ = 30º

According to the Bragg's law of diffraction,

For 1st secondary maxima,

Putting the value of a from Eqn. (i) to Eqn. (ii), we get

A: I

B: I

C: I

D: 3I

Ans:

Solution:

Given,

Maximum intensity = Io

Distance between the slits, d = 5λ

Distance of screen from the slit, D = 10d

Using the formula,

Path difference,

Here,

D = 10d = 50λ

So,

Corresponding phase difference will be

A: E/C

B: E/C

C: 2E/C

D: 2E/C

Ans:

Solution:

The radiation energy is given by

Initial momentum of the radiation is

The reflected momentum is

So, the change in momentum of light is

Thus, the momentum transferred to the surface is

A: 0.02 mm

B: 0.2 mm

C: 0.1 mm

D: 0.5 mm

Ans:

Solution:

In a double slit experiment, the two slits are 1 mm apart.

d = 1 mm = 10

The screen is placed at a distance D = 1 m away. Monochromatic light of wave length

λ = 500 nm = 5 x 10

The distance between two successive maxima or two successive minima is

**Q.13. ****For a parallel beam of monochromatic light of wavelength ‘λ’, diffraction is produced by a single slit whose width ‘a’ is of the order of the wavelength of the light. If ‘D’ is the distance of the screen from the slit, the width of the central maxima will be : (2015)A: 2Da/λB: 2Dλ/aC: Dλ/aD: Da/λAns: BSolution:**

For a parallel beam of monochromatic light of wavelength λ, diffraction is produced by a single slit whose width 'a' is of the order of the wavelength we have

where θ is the angle subtended by the first minima and the central maxima at the slit.

If x is the width of the central maxima, we have

where D is the distance of the screen from the slit**Q.14. ****A beam of light of λ = 600 nm from a distant source falls on a single slit 1mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between first dark fringes on either side of the central bright fringe is : (2014)A: 2.4 cmB: 2.4 mmC: 1.2 cmD: 1.2 mmAns: **B

Solution:

From the above figure,

For small θ and when θ is counted in rad, tan ≈ θ

So,**Q.15.**** In the Young’s double−slit experiment, the intensity of light at a point on the screen where the path difference is λ is K, (λ being the wave length of light used). The intensity at a point where the path difference is λ/4, will be : (2014)A: K/2B: ZeroC: KD: K/4Ans: ASolution:**

I = 4I

4I

∵ δ = 2π if path different = λ

Phase difference when path difference = Δ/4 is equal to

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