Question

The 546.1 nm line in mercury is measured at an angle of 73.2° in the third-order spectrum of a diffraction grating. Calculate the number of lines per centimeter for the grating.

Answer 1

Answer:

There are 5847.95 lines per cm for the grating.

Explanation:

Given that,

Wavelength of mercury line, $\lambda=546.1\ nm=546.1\times 10^{-9}\ m$

Angle in the third order spectrum, $\theta=73.2^{\circ}$

Using the grating equation, we get :

$d\ \sin\theta=m\lambda$

Here, m = 3

$d=\dfrac{m\lambda}{\sin\theta}\\\\d=\dfrac{3\times 546.1\times 10^{-9}}{\sin(73.2)}\\\\d=1.71\times 10^{-6}\ m$

Let there are N lines for the grating. So,

$N=\dfrac{1}{d}\\\\N=\dfrac{1}{1.71\times 10^{-6}\ m}\\\\N=\dfrac{1}{1.71\times 10^{-6}\times 10^2\ cm}\\\\N=\dfrac{1}{1.71\times 10^{-4}\ cm}\\\\N=5847.95\ \text{lines}/\text{cm}$

So, there are 5847.95 lines per cm for the grating.

Answer 2

Answer:

5841.42 lines per cm

Explanation:

wavelength, λ = 546.1 nm = 546 x 10^-9 m

Angle, θ = 73.2°

order, m = 3

Let d is the separation between the slits

d x Sinθ = m x λ

d x Sin 73.2 = 3 x 546.1 x 10^-9

d x 0.957 = 1638.3 x 10^-9

d = 1.7119 x 10^-6 m

d = 1.7119 x 10^-4 cm

Let N be the number of lines per cm.

So, N = 1 /d

$N = \frac{1}{1.7119 \times 10^{-4}}$

N = 5841.42 lines per cm