Question

Light of wavelength 6328 Å is incident normally
on a slit having width of 0.2 mm. Width from first
minima to first minima of diffraction pattern on a
screen 9m away will be:
(1) 0.36° (2) 0.18° (3) 0.72 (4) 0.090

​

Answer 1

Correct Question:

Light of wavelength 6328 Å is incident normally on a slit having width of 0.2 mm. Width from first minima to first maxima of diffraction pattern on a screen 9m away will be:

(1) 0.36°

(2) 0.18°

(3) 0.72°

$\sf \checkmark{(4)} \ 0.09\degree$

Given:

Wavelength of incident light (λ) = 6328 Å = $\sf 6328 \times 10^{-10}$ m

Width of slit (a) = 0.2 mm = $\sf 2 \times 10^{-4}$ m

Distance between screen and slit (D) = 9 m

To Find:

Width from first minima to first maxima of diffraction pattern on a screen

Answer:

For a bright point:

$\bf \theta_n = \dfrac{(2n + 1)\lambda}{2a}$

$\rm For \: 1^{st} \: bright \: point : \\ \rm \implies {\theta_1}_{bright} = \dfrac{(2 \times 1 + 1)\lambda}{2a} \\ \\ \rm \implies {\theta_1}_{bright} = \dfrac{3\lambda}{2a}$

For a dark point:

$\bf \theta_n = \dfrac{n \lambda}{a}$

$\rm For \: 1^{st} \: dark \: point : \\ \rm \implies {\theta_1}_{dark} = \dfrac{1 \times \lambda}{a} \\ \\ \rm \implies {\theta_1}_{bright} = \dfrac{\lambda}{a}$

Width from $\sf 1^{st}$ minima to $\sf 1^{st}$ maxima = $\rm {\theta_1}_{bright} - {\theta_1}_{dark}$

$\rm \implies Width = \dfrac{3\lambda}{2a} - \dfrac{\lambda}{a} \\ \\ \rm \implies Width = \dfrac{3\lambda - 2 \lambda}{2a} \\ \\ \rm \implies Width = \dfrac{\lambda}{2a} \\ \\ \rm \implies Width = \dfrac{6328 \times {10}^{ - 10} }{2 \times 2 \times {10}^{ - 4} } \\ \\ \rm \implies Width = 1582 \times {10}^{ - 6} \: rad$

In degree;

$\rm \implies Width = 1582 \times {10}^{ - 6} \times \dfrac{180}{\pi} \\ \\ \rm \implies Width = 0.09 \degree$

$\therefore$ Width from first minima to first maxima of diffraction pattern on a screen will be 0.09°

i.e. $\boxed{\mathfrak{(4) \ 0.09 \degree}}$