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: Approach
Use the ratio of the two widths.

Step 2: Ratio
At the same $\lambda$, $a\sin\theta$ is the same, so $\dfrac{a_2}{a_1}=\dfrac{\sin30^\circ}{\sin45^\circ}=\dfrac{1/2}{1/\sqrt2}=\dfrac{1}{\sqrt2}$.

Step 3: Widths
With $a_1=3\lambda$ we get $a_2=\dfrac{3\lambda}{\sqrt2}$. The change is $3\lambda\left(1-\dfrac1{\sqrt2}\right)=3\lambda\dfrac{\sqrt2-1}{\sqrt2}$. 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). \[ \boxed{3\left(\frac{\sqrt2-1}{\sqrt2}\right)\lambda} \]