Question

Light of wavelength 530.00 nm is incident normally on a diffraction grating, and the first‑order maximum is observed to be 33.0∘ from the normal. How many slits per millimeter are marked on the grating?

Answer 1

Answer:

1028 slits/mm

Explanation:

We are given that

Wavelength of light, $\lambda=530nm=530\times 10^{-9} m$

1nm=$10^{-9} m$

$\theta=33^{\circ}$

n=1

We have to find the number of slits per mm are marked on the grating.

We know that

$dsin\theta=n\lambda$

Using the formula

$dsin33^{\circ}=1\times 530\times 10^{-9}$

$d=\frac{530\times 10^{-9}}{sin33^{\circ}}$

$d=9.731\times 10^{-7} m$

1m=$10^{3}mm$

$d=9.731\times 10^{-7}\times 10^3$mm

$d=0.0009731mm$

Number of slits=$\frac{1}{d}$

Number of slits=$\frac{1}{0.0009731}$/mm

Number of slits=1028/mm

Hence, 1028 slits/mm are marked on the grating.

Answer 2

Answer:

1027.6 lines per mm.

Explanation:

wavelength = 530 nm

order, m= 1

Angle = 33 degree

Let the slits per mm is 1/d.

So,

$m \lambda = d sin A\\\\1\times 530\times 10^{-6} = d sin 33\\\\\frac{1}{d} = 1027.6 lines per mm$