MHTCET 16 May 2024 Shift 1 Paper

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Question : 18 of 150

Marks: +1, -0

The angular separation of the central maximum in the Fraunhofer diffraction pattern is measured. The slit is illuminated by the light of wavelength

6000 \text{\AA}^{o}

. If the slit is illuminated by light of another wavelength, the angular separation decreases by 20\%. The wavelength of light used is

$6400 \stackrel{o}{A}$
$5600 \stackrel{o}{A}$
$4800 \stackrel{o}{A}$
$4400 \stackrel{o}{A}$
The angular width of the central maximum in a Fraunhofer diffraction pattern is given by:
$\theta=\frac{2 \lambda}{a}$
This implies that:
$\theta \propto \lambda$
Therefore, if the angle changes when a different wavelength is used, we get:
$\frac{\theta_{2}}{\theta_{1}}=\frac{\lambda_{2}}{\lambda_{1}}$
Given that the angular separation decreases by 20\% when a new wavelength is used, we have:
$\theta_{2}=\theta_{1}-\frac{20}{100} \theta_{1}=0.8 \theta_{1}$
Thus:
$\lambda_{2}=0.8 \lambda_{1}=0.8 \times 6000=4800 \stackrel{\circ}{A}$
Therefore, the wavelength of the light used is $4800 \stackrel{o}{A}$ .

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