Question

When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.0 ∘ from the central maximum. How many additional pairs of bright spots are there beyond the first bright spots?

A converging lens 6.90 cm in diameter has a focal length of 310 mm If the resolution is diffraction limited, how far away can an object be if points on it transversely 4.00 mm apart are to be resolved (according to Rayleigh's criterion) by means of light of wavelength 580 nm ?

Using Rayleigh's criterion, what is the smallest separation between two pointlike objects that a person could clearly resolve at his near point of 21.0 cm with a pupil diameter of 2.15 mm ? Assume that the light has a wavelength of 504 nm .

Answer 1

(i) The line density of a grating which will be given by -

N = sin $\theta$ / $\lambda$

where, $\lambda$ = wavelength of laser light = 632.8 x 10^-9 m

$\theta$ = first bright spots occur at an angle = 17⁰

then, we get

N = sin 17⁰ / (632.8 x 10^-9 m)

N = [(0.2923) / (632.8 x 10^-9 m)]

N = 4.62 x 10⁵ lines/m

converting lines/m into lines/cm :

N = 4.62 x 10³ lines/cm

How many additional pairs of bright spots are there beyond the first bright spots?

m = 1 / $\lambda$ N

m = 1 / [(632.8 x 10^-9 m) (4.62 x 10⁵ lines/m)]

m = 1 / (0.2923536)

m = 3.42

Then, we get

n = 3 + (first bright spot)

n = 4 additional

(ii) According to Rayleigh criterion, we have

Sin $\theta$ = (1.22) ($\lambda$ / D)

where, $\lambda$ = wavelength of light = 580 x 10^-9 m

D = diameter of a converging lens = 6.90 x 10^-2 m

then, we get

Sin $\theta$ = (1.22) [(580 x 10^-9 m) / (6.90 x 10^-2 m)]

Sin $\theta$ = 1.02 x 10^-5 rad

Angle between the two points at the diffraction limit.

We know that, Sin $\theta$ = (L / d)

(1.025 x 10^-5 rad) = (4 x 10^-3 m) / d

d = [(4 x 10^-3 m) / (1.025 x 10^-5 rad)]

d = 392.1 m

(iii) According to Rayleigh criterion, we have

Sin $\theta$ = (1.22) ($\lambda$ / D)

where, $\lambda$ = wavelength of light = 504 x 10^-9 m

D = diameter of a pupil = 2.15 x 10^-3 m

then, we get

Sin $\theta$ = (1.22) [(504 x 10^-9 m) / (2.15 x 10^-3 m)]

Sin $\theta$ = 2.86 x 10^-4 rad

The smallest separation distance between two point objects by human eye is given by -

d = x Sin $\theta$

where, x = distance of the near point = 21 x 10^-2 m

then, we get

d = [(21 x 10^-2 m) (2.86 x 10^-4 rad)]

d = 6.0 x 10^-5 m