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Question: A screen is placed \(50cm\)from a single slit which is illuminated with light of wavelength\(6000{{A...

A screen is placed 50cm50cmfrom a single slit which is illuminated with light of wavelength6000A06000{{A}^{0}}. If the distance between the first and third minima in the diffraction pattern is3.0mm3.0mm, the width of the slit is
(A) 0.1mm0.1mm
(B) 0.2mm0.2mm
(C) 0.3mm0.3mm
(D) 0.4mm0.4mm

Explanation

Solution

Hint : In this question, we have to find the width of the slit. We have given the distance between the first and third minima in the diffraction pattern in the single-slit diffraction and the wavelength of the light so we use this formula to find out the width of the slit is given x2−x1=(2λDd){{x}_{2}}-{{x}_{1}}=(\dfrac{2\lambda D}{d}).

Complete step by step answer:
Given: We have given the following terms in this question are
x2−x1=3mm{{x}_{2}}-{{x}_{1}}=3mm Here we convert mmmm into mm then we get,
Δx=3×10−3m\Delta x=3\times {{10}^{-3}}m
D=50cmD=50cm Here we convert cmcm into mm then we get,
D=0.5mD=0.5m
λ=6000A0\lambda =6000{{A}^{0}} Here we convert A0{{A}^{0}}into mmthen we get,
λ=6×10−10m\lambda =6\times {{10}^{-10}}m
Here is the formula for single slit diffraction is given as,
x2−x1=(3λ−λ)Dd{{x}_{2}}-{{x}_{1}}=\dfrac{(3\lambda -\lambda )D}{d}
We simplify this formula as per the given conditions,
Δx=2λDd\Delta x=\dfrac{2\lambda D}{d}
Here,
Δx\Delta x= The distance between the first and third minima in the diffraction pattern.
DD= The distance of a screen from a single slit.
λ\lambda = Wavelength of light.
dd= Width of the slit.
Now we substitute the given values in the formula,
d=2λDΔxd=\dfrac{2\lambda D}{\Delta x}
After substituting the values equation is written as,
d=2×6000×10−10×0.53×10−3d=\dfrac{2\times 6000\times {{10}^{-10}}\times 0.5}{3\times {{10}^{-3}}}
After simplifying this we get,
d=2×10−4md=2\times {{10}^{-4}}m
We have to calculate the width of the slit mmmm.
d=0.2mmd=0.2mm

So option B is the correct answer.

Note: To solve this question we have to study the diffraction of light and the single slit diffraction experiment. In this experiment, we can observe the bending phenomenon of light or diffraction that causes light from a coherent source to interfere with itself and produce a distinctive pattern on the screen called the diffraction pattern. Using this experiment we can solve questions based on the single slit experiment easily.