Question

Find the ratio of intensity at $1^{st}$ maximum to intensity at $2^{nd}$ maximum in a single slit diffraction experiment.

• A. $2.78$
• B. $1$
• C. $2.56$
• D. $3.82$

Answer 1

The correct option is A $2.78$

Intensity at a point on the screen is given by

$I=I_0{\left(\frac{\sin \beta }{\beta }\right)}^2$

Where, $I_0$ Intensity at central maximum.

$\beta =\frac{\pi }{\lambda }\left(b\sin \theta \right)$

The angular position of $1^{st}$ maximum is

$\sin {\theta }_1=\frac{3\lambda }{2b}$

$\Rightarrow {\beta }_1=\frac{3\pi }{2}$

$\Rightarrow I_1=I_0{\left(\frac{\sin \left(3\pi /2\right)}{3\pi /2}\right)}^2$

$\Rightarrow I_1=I_0\left(\frac{49}{1089}\right)$

The angular position of $2^{nd}$ maximum is

$\sin {\theta }_2=\frac{5\lambda }{2b}$

$\Rightarrow {\beta }_2=\frac{5\pi }{2}$

$\Rightarrow I=I_0{\left(\frac{\sin \left(5\pi /2\right)}{5\pi /2}\right)}^2$

$\Rightarrow I_2=I_0\left(\frac{49}{3025}\right)$

$\therefore \frac{I_1}{I_0}=\frac{3025}{1089}=2.78$

Hence, option $\left(\text{A}\right)$ is correct.