Question

-
0
A parallel beam of wave
length 1 = 4500 A passes
through a long slit of width
2x104m. The angular
divergence for which most of
the light is diffracted is
(inx10-* radian)​

Answer 1

Note: The width of the slit is$\[2 \times {10^{ - 4}}m.\]$

Given,

The length of the parallel beam$\[ = 4500\]$

The width of the slit$\[ = 2 \times {10^{ - 4}}\]$

Solution,

Know that most of the light is diffracted between the two first-order minima.

Therefore, the required angular divergence is

$\[\begin{array}{*{20}{l}}{ = 2\theta }\\{ = 2\sin \theta }\\{ = \frac{{2\lambda }}{d}}\end{array}\]$

Apply values.

$\[\begin{array}{*{20}{l}}{ = \frac{{2\left( {450 \times {{10}^{ - 9}}} \right)}}{{2 \times {{10}^{ - 4}}}}}\\{ = 4.5 \times {{10}^{ - 3}}}\end{array}\]$

Hence, the angular divergence for which most of the light is diffracted is $\[4.5 \times {10^{ - 3}}\]$radians.