Question

In a diffraction pattern due to single slit of width 'a', the first minimum is observed at an angle of 30° when the light of wavelength $5400\text{ \AA}$ is incident on the slit. The first secondary maximum is observed at an angle of ($\sin 30^\circ = \frac{1}{2}$)

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

Hint:

Logic Tip: The ratio of the sine of the angles for the 1st maximum to the 1st minimum is simply the ratio of their path differences: $\frac{1.5\lambda}{1.0\lambda} = 1.5$. Thus, $\sin \theta_{max} = 1.5 \times \sin(30^\circ) = 1.5 \times 0.5 = 0.75 = \frac{3}{4}$.

Answer 1

The Correct Option is A

Concept:
For a single slit diffraction pattern, the angular position $\theta$ for the $n$-th minimum is given by $a \sin \theta = n\lambda$. The angular position $\theta$ for the $n$-th secondary maximum is approximately given by $a \sin \theta = (2n+1)\frac{\lambda}{2}$.
Step 1: Use the condition for the first minimum.
For the first minimum, $n=1$. $$a \sin \theta_{min} = 1 \cdot \lambda$$ We are given $\theta_{min} = 30^\circ$, so: $$a \sin(30^\circ) = \lambda$$ $$a \left(\frac{1}{2}\right) = \lambda \implies a = 2\lambda$$
Step 2: Set up the condition for the first secondary maximum.
For the first secondary maximum, $n=1$. $$a \sin \theta_{max} = (2(1) + 1)\frac{\lambda}{2}$$ $$a \sin \theta_{max} = \frac{3\lambda}{2}$$
Step 3: Solve for the angle of the maximum.
Substitute $a = 2\lambda$ into the maximum equation: $$(2\lambda) \sin \theta_{max} = \frac{3\lambda}{2}$$ Divide both sides by $\lambda$: $$2 \sin \theta_{max} = \frac{3}{2}$$ $$\sin \theta_{max} = \frac{3}{4}$$ $$\theta_{max} = \sin^{-1}\left(\frac{3}{4}\right)$$