Question

A 0.52 mm-diameter hole is illuminated by light of wavelength 510 nm.

What is the width of the central maximum on a screen 2.2 m behind the slit?

Answer 1

Concepts and reason

The concept required to solve the given problem is of diffraction with circular aperture.

Initially, simplify the equation of central width by substituting the equation of angular separation in it. Then substitute the given values to find the value of the width of the central maximum on the screen.

Fundamentals

The expression for the width of the central maximum is given by,

$x = 2L\theta$ …… (1)

Here, x is the width of the central maximum, L is the screen distance, and $\theta$ is the angular separation.

The angular separation is given by,

$\theta = \frac{{1.22\lambda }}{D}$ …… (2)

Here, $\theta$ is the angular separation, $\lambda$ is the wavelength, and D is the diameter of the aperture.

The expression for the width of the central maximum is given by,

$x = 2L\theta$

Substitute equation (2) in the above equation.

$x = 2L\left( {\frac{{1.22\lambda }}{D}} \right)$ …… (3)

The expression for the width of the central maximum is,

$x = 2L\left( {\frac{{1.22\lambda }}{D}} \right)$

Substitute 2.2 m for L, 510 nm for $\lambda$ and 0.52 mm for D in the above equation.

$\begin{array}{c}\\x = 2\left( {2.2\;{\rm{m}}} \right)\left( {\frac{{1.22\left( {510\;{\rm{nm}}} \right)}}{{0.52\;{\rm{mm}}}}} \right)\\\\ = 2\left( {2.2\;{\rm{m}}} \right)\left( {\frac{{1.22\left( {510\;{\rm{nm}}\left( {\frac{{1\;{\rm{m}}}}{{{{10}^9}\;{\rm{nm}}}}} \right)} \right)}}{{0.52\;{\rm{mm}}\;\left( {\frac{{1\;{\rm{m}}}}{{{{10}^3}\;{\rm{mm}}}}} \right){\rm{m}}}}} \right)\\\\ = 5.26 \times {10^{ - 3}}\;{\rm{m}}\\\\{\rm{ = }}5.26\;{\rm{mm}}\\\end{array}$

The width of the central maximum is 5.26 mm.

Ans:

The width of the central maximum is 5.26 mm.