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Question: A slit of width \(a\) is illuminated by white light. For red light \(\left( {\lambda = 6500\mathop {...

A slit of width aa is illuminated by white light. For red light (λ=6500A0  )\left( {\lambda = 6500\mathop {\rm{A}}\limits^0 \;} \right). The first minima is obtained at θ=30\theta = 30^\circ . Then the value of aa will be
(A) 3250A03250\mathop {\rm{A}}\limits^0
(B) 6.5×104  mm6.5 \times {10^{ - 4}}\;{\rm{mm}}
(C) 1.24  microns1.24\;{\rm{microns}}
(D)2.6×104  cm2.6 \times {10^{ - 4}}\;{\rm{cm}}

Explanation

Solution

In a slit experiment, the formula of the slit width is a=λθa = \dfrac{\lambda }{\theta }. Where, aa is the width of the slit, θ\theta is the angle of the beam and λ\lambda is the wavelength of the light source.
The wavelength of the light is different for different colors that depend on the spectrum of light.
According to the formula of width of the slit, the width of the slit increases when the angle of beam decreases, and the width of the slit decreases when angle of beam increases.
In the slit experiment, the width of the slit depends on the angle of beam and the wavelength of the light source.

Complete step by step answer:
Given data:

The wavelength of the red light is λ=6500A0\lambda = 6500\mathop {\rm{A}}\limits^0
The first minima is obtained at an angle θ=30\theta = 30^\circ .

Convert the wavelength of the red light from angstrom to meter.

\lambda = 6500\mathop {\rm{A}}\limits^0 \left( {\dfrac{{{{10}^{ - 10}}\;{\rm{m}}}}{{1\;\mathop {\rm{A}}\limits^0 \;}}} \right)\\\ = 6500 \times {10^{ - 10}}\;{\rm{m}} \end{array}$$ So, the wavelength of the red light in meters is $$6500 \times {10^{ - 10}}\;{\rm{m}}$$. Convert the angle $\theta $ from degree to radian. $$\begin{array}{c} \theta = 30^\circ \\\ = \dfrac{\pi }{{180^\circ }} \times 30^\circ \;\;\\\ = \dfrac{\pi }{6} \end{array}$$ So, the angle of the beam in the radian is $$\dfrac{\pi }{6}$$. Substitute $$6500 \times {10^{ - 10}}\;{\rm{m}}$$ for $$\lambda $$ and $$\dfrac{\pi }{6}$$ for $$\lambda $$ In the equation $a = \dfrac{\lambda }{\theta }$ to find the width of the slit. $\begin{array}{c} a = \dfrac{\lambda }{\theta }\\\ = \dfrac{{6500 \times {{10}^{ - 10}}\;{\rm{m}}}}{{\dfrac{\pi }{6}}}\\\ \approx 1.24 \times {10^{ - 6}}\;{\rm{m}} \end{array}$ Convert the width of the slit from meter to microns. $$\begin{array}{c} a = 1.24 \times {10^{ - 6}}\;{\rm{m}}\left( {\dfrac{{{{10}^{ - 6}}\;{\rm{microns}}}}{{1\;{\rm{m}}}}} \right)\\\ = 1.24\;{\rm{microns}} \end{array}$$ Therefore, the width of the slit is $$1.24\;{\rm{microns}}$$ **So, the correct answer is “Option C”.** **Note:** Make sure that the conversion of each unit is in proper form to find the exact solution. To find the width of the slit it is necessary to first convert the angle of beam into radian and then put the value in the formula.