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UPSEE 2018 Solved Paper

Question : 5 of 150

Marks: +1, -0

The diffraction pattern of a single slit is shown in figure. The point at which the path difference of the extreme rays is two wavelengths is

point 1
point 2
point 4
point 5

Validate

Solution:

Diffraction of light at a single slit, the pattern obtained consists of a central bright band having alternate dark and weak bands of decreasing intensity on both sides.

For nth minima (secondary),

\sin \theta_{n} = \frac{n \lambda}{a}

For nth maxima (secondary),

\sin \theta_{n} = (2 n + 1) \frac{\lambda}{2 a}

Thus, point 5 corresponds to path difference of extreme rays equal to

2 \lambda

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