MHT CET · Physics · Wave Optics

In a Fraunhofer diffraction at a single slit of width \(d\) and incident light of wavelength \(5500 \mathrm{~A}\), the firs minimum is observed at an angle \(30^{\circ}\). The first secondary maxima are observed at an angle \(\theta=\)

A \(\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
B \(\sin ^{-1}\left(\frac{1}{4}\right)\)
C \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
D \(\sin ^{-1}\left(\frac{3}{4}\right)\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\sin ^{-1}\left(\frac{3}{4}\right)\)

Step-by-step Solution

Detailed explanation

Slit width \(=d\)
\(\lambda=5500\)Å \(=5.5 \times 10^{-7} \mathrm{~m}, \theta_1=30^{\circ}\)
For first secondary minima, \(d \sin \theta_1=\lambda\)
\(d=\frac{\lambda}{\sin \theta_1}=\frac{5.5 \times 10^{-7}}{\sin 30^{\circ}}=11 \times 10^{-7}\)
For first secondary maxima, \(d \sin \theta_1=\frac{3 \lambda}{2}\)
\(\begin{aligned} & \Rightarrow \sin \theta_1=\frac{3 \lambda}{2 d}=\frac{5.5 \times 10^{-7}}{2 \times 11 \times 10^{-7}}=\frac{3}{4} \\ & \theta_1=\sin ^{-1}\left(\frac{3}{4}\right)\end{aligned}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app