    Question

# A parallel beam of monochromatic light of wavelength $5000\stackrel{\circ }{A}$ is incident on a single narrow slit of width $0.001mm$. The light is focused by a convex lens on a screen placed on a focal plane. The first minimum will be formed for the angle of diffraction equal to

A

$50°$

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B

$15°$

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C

$20°$

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D

$30°$

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Solution

## The correct option is D $30°$Step 1: Given dataWavelength of light, $\lambda =5000A°$Slit width, $d=0.001mm$Step 2: To findThe angle of diffraction in which the first minimum will be formed.Step 3: Formula used$d\mathrm{sin}\theta =n\lambda$where $d$ is the slit width, $\lambda$ is the wavelength of light and $n$ is the nth minima.Step 4: Calculation for the angle:Here since the first minimum is considered, $n=1$$d\mathrm{sin}\theta =n\lambda$Substituting the values,$d\mathrm{sin}\theta =n\lambda \phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\frac{n\lambda }{d}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\frac{1×5000×{10}^{-10}}{0.001×{10}^{-3}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\theta ={\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}⇒\theta =30°$The first minimum will be formed for the angle of diffraction equal to $30°$Option D is correct  Suggest Corrections  0      Similar questions  Explore more