Question

In the diffraction pattern, the first maximum is at \(30^{\circ}\), when a monochromatic light of wavelength \(λ\) 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 \((sin30^{\circ} = \frac{1}{2},sin45^{\circ} = \frac{1}{\sqrt{2}})\)

• A. \(3(\frac{\sqrt{2}-1}{\sqrt{2}})λ\)
• B. \(\frac{3}{\sqrt{2}}λ\)
• C. \(3\sqrt{2}λ\)
• D. \(3(\frac{\sqrt{2}+1}{\sqrt{2}})λ\)

Hint:

The first secondary maximum satisfies a sin(theta) = 3 lambda over 2.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For a single slit, the first secondary maximum (the first maximum beside the centre) occurs at
\[ a\sin\theta=\frac{3\lambda}{2} \]

Step 2: Slit width in each case
\[ a_1=\frac{3\lambda}{2\sin30^\circ}=3\lambda,\qquad a_2=\frac{3\lambda}{2\sin45^\circ}=\frac{3\lambda}{\sqrt2} \]

Step 3: Change
\[ a_1-a_2=3\lambda-\frac{3\lambda}{\sqrt2}=3\lambda\left(\frac{\sqrt2-1}{\sqrt2}\right) \]
So the slit width is reduced by this amount. This is option (A).

Final Answer:
The widths are 3 lambda and 3 lambda over root 2, so the change is 3(root 2 - 1)/root 2 times lambda, option (A). \[ \boxed{3\left(\frac{\sqrt2-1}{\sqrt2}\right)\lambda} \]