Question

Monochromatic light of wavelength 400 nm falls normally on a slit of width 0.1 mm. Find the angular positions of first secondary maximum and second minimum in the diffraction pattern.

a)20 ' , 27 '
b)27 ' , 20 '

Answer 1

Given:

Light of wavelength 400 nm falls normally on a slit of width 0.1 mm

To find:

Angular width of

• 1st secondary maxima

• Second minima

Calculation:

The general formula for angular width for any order maxima or minima (except the central maxima) is :

$\boxed{ \sf{ \phi = \dfrac{ \lambda}{ d} }}$

$\lambda$ refers to wavelength and d refers to the inter-slit distance.

For the 1st secondary maxima:

$\therefore \: \phi1 = \dfrac{ \lambda}{d}$

$= > \: \phi1 = \dfrac{400 \times {10}^{- 9} }{0.1 \times {10}^{ - 3} }$

$= > \: \phi1 = \dfrac{400 \times {10}^{- 9} }{{10}^{ - 4} }$

$= > \: \phi1 = 400 \times {10}^{- 5}$

$= > \: \phi1 = 4 \times {10}^{- 3} \: rad$

$= > \: \phi1 = 13.75 \: minutes$

For the 2nd minima:

$\therefore \: \phi2= \dfrac{ \lambda}{d}$

$= > \: \phi2= \dfrac{400 \times {10}^{- 9} }{0.1 \times {10}^{ - 3} }$

$= > \: \phi2 = \dfrac{400 \times {10}^{- 9} }{{10}^{ - 4} }$

$= > \: \phi2= 400 \times {10}^{- 5}$

$= > \: \phi2= 4 \times {10}^{- 3} \: rad$

$= > \: \phi2= 13.75 \: minutes$

So, final answer is :

$\boxed{\sf{\phi1 = \phi2 = 13.75 '}}$