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Question: Light of wavelength 5000Å is incident normally on a slit of width \[2.5 \times 10_{}^{ - 4}\] cm. Th...

Light of wavelength 5000Å is incident normally on a slit of width 2.5×10−42.5 \times 10_{}^{ - 4} cm. The angular position of second minimum from the central maximum is
A. sin⁡−1(15){\sin ^{ - 1}}\left( {\dfrac{1}{5}} \right)
B. sin⁡−1(25){\sin ^{ - 1}}\left( {\dfrac{2}{5}} \right)
C. (π3)\left( {\dfrac{\pi }{3}} \right)
D. (π6)\left( {\dfrac{\pi }{6}} \right)
E. (π4)\left( {\dfrac{\pi }{4}} \right)

Explanation

Solution

Consider a single slit diffraction. The point in that diffraction pattern where the secondary waves reinforce each other. In that particular point the intensity is maximum and it is known as Central maximum.
Formula used:
sinθ = 2λa {\text{sin}}\theta {\text{ = }}\dfrac{{2\lambda }}{a}{\text{ }}
Where,
λ\lambda is the wavelength

Complete step-by-step answer:
Let us consider the wavelength of the given light
Given that light of wavelength 5000Å is incident normally on a slit of width 2.5×10−42.5 \times 10_{}^{ - 4} cm.
Hence wavelength,
λ=5000Å= 5000×10−105000 \times 10_{}^{ - 10} m
Let us now consider the width.
width of slit, a= 2.5×10−42.5 \times 10_{}^{ - 4} cm= 2.5×10−62.5 \times 10_{}^{ - 6}m.
We can now use the formula for angular position of second minimum from the central maximum,
sinθ = 2λa    {\text{sin}}\theta {\text{ = }}\dfrac{{2\lambda }}{a}{\text{ }} \\\ \\\
We can now substitute the values of λ\lambda in the above given formula.,
⇒sinθ = 2×5000×10−102.5×10−6 \Rightarrow {\text{sin}}\theta {\text{ = }}\dfrac{{2 \times 5000 \times 10_{}^{ - 10}}}{{2.5 \times 10_{}^{ - 6}}}{\text{ }}
⇒sinθ = 10000×10−102.5×10−6\Rightarrow {\text{sin}}\theta {\text{ = }}\dfrac{{10000 \times 10_{}^{ - 10}}}{{2.5 \times 10_{}^{ - 6}}}
⇒sinθ = 1025\Rightarrow {\text{sin}}\theta {\text{ = }}\dfrac{{10}}{{25}}
⇒sin⁡θ=25\Rightarrow \sin \theta = \dfrac{2}{5}
Therefore, the angular position of the second minimum from the second maximum is sin⁡−1(25){\sin ^{ - 1}}\left( {\dfrac{2}{5}} \right)
hence, option (B) sin⁡−1(25){\sin ^{ - 1}}\left( {\dfrac{2}{5}} \right) is the required correct answer.
Additional information:
The diffraction pattern due to a single slit consists of a central bright maximum followed by alternate secondary minima and maxima on both sides.
The width of the central maximum is proportional to the wavelength of light. Hence with red light, the width of the central maximum is more than the violet light.
When the slit is narrow the width of the central maximum is more.
If the width of the slit is large, sin θ is small and hence θ is large. This results in a distinct diffraction maximum and minimum on both the sides of central maximum point.
Note:
The angular half width of the central maximum is given by θ=λa\theta = \dfrac{\lambda }{a} and angular width of central maximum is given by2θ=2λa2\theta = \dfrac{{2\lambda }}{a} , if θ is small.
It is very important to consider if the given point is of maximum or minimum.