Question

monochromatic light of wavelength 4300 A falls on a slit of width a . for what value of a does the first maximum at 30
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Answer 1

Answer:

Width of the slit in single elit experiment is given as

$a = 1.29 \mu m$

Explanation:

As we know that light of given wavelength when pass through a single slit then due to diffraction of light we get the pattern of maximum and minimum intensity on the screen

For maximum intensity on screen we know that

$a sin\theta = \frac{2N + 1}{2}\lambda$

for 1st maximum on the screen we have N = 1

so we will have

$a sin\theta = \frac{3}{2}\lambda$

now we know that

$\theta = 30^o$

$a sin30 = \frac{3}{2} (4300)$

$a = 12900 \times 10^{-10} m$

$a = 1.29 \mu m$

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Topic : Single slit diffraction

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