Question

The angular deviation of 5th
order dark fringe is 12° in a
single slit experiment. If the
width of the slit is 9 um then the
wave length of the incident
light is​

Answer 1

Explanation:

1. 2dsin°=m¥
2. solve with this equation......equation

Answer 2

Given:

The angular deviation of 5thorder dark fringe is 12° in a single slit experiment, the

width of the slit is 9 um.

To find:

Wave length of the incident light.

Calculation:

The general expression of angular deviation of any fringe (except the central fringe) in single slit diffraction is given as:

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

Putting available values:

$= >12 \times \dfrac{\pi}{180} = \dfrac{ \lambda}{9 \times {10}^{ - 9} }$

$= > \dfrac{\pi}{15} = \dfrac{ \lambda}{9 \times {10}^{ - 9} }$

$= > \: \lambda = \dfrac{9\pi}{15} \times {10}^{ - 9}$

$= > \: \lambda = 1.88 \times {10}^{ - 9}$

$= > \: \lambda = 18.8 \times {10}^{ - 10} \: m$

$= > \: \lambda = 18.8 \: {A}^{ \circ}$

So , final answer is:

$\boxed{\: \lambda = 18.8 \: {A}^{ \circ} }$