Q103 / 150 · MHT CET 2026 (20 April Shift 2) · Physics
In the diffraction pattern, the first maximum is at $30^\circ$, when a monochromatic light of wavelength $\lambda$ is incident on a slit of width $a$. For the same wavelength, if the slit width is changed, so that first maximum is at $45^\circ$. The slit width is changed by $\left(\sin 30^\circ = \dfrac{1}{2}, \sin 45^\circ = \dfrac{1}{\sqrt{2}}\right)$
- Correct answerA$3\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)\lambda$
- B$\dfrac{3}{\sqrt{2}}\lambda$
- C$3\sqrt{2}\lambda$
- D$3\left(\dfrac{\sqrt{2}+1}{\sqrt{2}}\right)\lambda$
Solution
The condition for the first secondary maximum in a single slit diffraction pattern is given by:
$a \sin \theta = \dfrac{3\lambda}{2}$
For the initial case, $\theta_1 = 30^\circ$:
$a_1 \sin 30^\circ = \dfrac{3\lambda}{2}$
$a_1 \left(\dfrac{1}{2}\right) = \dfrac{3\lambda}{2} \Rightarrow a_1 = 3\lambda$
For the final case, $\theta_2 = 45^\circ$:
$a_2 \sin 45^\circ = \dfrac{3\lambda}{2}$
$a_2 \left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{3\lambda}{2} \Rightarrow a_2 = \dfrac{3\sqrt{2}\lambda}{2} = \dfrac{3\lambda}{\sqrt{2}}$
The change in slit width is:
$\Delta a = a_1 - a_2 = 3\lambda - \dfrac{3\lambda}{\sqrt{2}} = 3\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)\lambda$
Answer: $3\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)\lambda$
$a \sin \theta = \dfrac{3\lambda}{2}$
For the initial case, $\theta_1 = 30^\circ$:
$a_1 \sin 30^\circ = \dfrac{3\lambda}{2}$
$a_1 \left(\dfrac{1}{2}\right) = \dfrac{3\lambda}{2} \Rightarrow a_1 = 3\lambda$
For the final case, $\theta_2 = 45^\circ$:
$a_2 \sin 45^\circ = \dfrac{3\lambda}{2}$
$a_2 \left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{3\lambda}{2} \Rightarrow a_2 = \dfrac{3\sqrt{2}\lambda}{2} = \dfrac{3\lambda}{\sqrt{2}}$
The change in slit width is:
$\Delta a = a_1 - a_2 = 3\lambda - \dfrac{3\lambda}{\sqrt{2}} = 3\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)\lambda$
Answer: $3\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)\lambda$
Hence, the correct answer is option: 'A'.