Chapter 27, Problem 1OQ

Consider a wave passing through a single slit. What happens to the width of the central maximum of its diffraction pattern as the slit is made half as wide? (a) It becomes one-fourth as wide. (b) It becomes one-half as wide. (c) Its width does not change. (d) It becomes twice as wide. (e) It becomes four times as wide.

To determine

The effect on width of central maximum on decreasing the size of slit width to half the initial value.

Answer to Problem 1OQ

Option (d)

Explanation of Solution

Write the condition for first order dark central maxima for diffraction produced by wave passing through the slit.

    $\sin {\theta }_m=\frac{\lambda }{a}$

Here, ${\theta }_m$ is the diffraction angle for dark fringe, $\lambda$ is the wavelength of light, and $a$ is the slit width.

Write the expression for tangent of ${\theta }_m$ on considering that ${\theta }_m$ has only very small values.

    $\sin {\theta }_m\approx \tan {\theta }_m$

Rewrite the first equation by substituting the above one.

  $\sin {\theta }_m=\frac{\lambda }{a}$        (I)

Express $\tan {\theta }_m$ in terms of separation between slits and screen.

    $\tan {\theta }_m=\frac{y_{\text{dark}}}{L}$

Here, $y_{\text{dark}}$ is the width of central maximum and $L$ is the separation between slit and screen.

Rewrite the above expression by substituting equation (I).

    $\frac{\lambda }{a}=\frac{y_{\text{dark}}}{L}$

Rewrite the expression for $y_{\text{dark}}$.

    $y_{\text{dark}}=\frac{L\lambda }{a}$

Conclusion:

Rewrite the above equation by replacing $a$ by $\frac{a}{2}$.

    $\begin{array}{c}{y^{\prime }}_{\text{dark}}=\frac{L\lambda }{\left(\frac{a}{2}\right)}\\ =2y_{\text{dark}}\end{array}$

It is found that as slit width reduces to half the initial value, width of central maximum will increases by two times. Thus, option (d) is correct.

${y^{\prime }}_{\text{dark}}$ to become one by fourth, $a$ has to increase by four times. Thus, option (a) is incorrect.

${y^{\prime }}_{\text{dark}}$ to become one by two, $a$ has to increase by two times. Thus, option (b) is incorrect.

${y^{\prime }}_{\text{dark}}$ to remain unchanged, $a$ also has to remain unchanged. Thus, option (c) is incorrect.

${y^{\prime }}_{\text{dark}}$ to become 4 times, $a$ has to decrease by four times. Thus, option (e) is incorrect.