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प्रश्न
White coherent light (400 nm-700 nm) is sent through the slits of a Young's double slit experiment (see the following figure). The separation between the slits is 0⋅5 mm and the screen is 50 cm away from the slits. There is a hole in the screen at a point 1⋅0 mm away (along the width of the fringes) from the central line. (a) Which wavelength(s) will be absent in the light coming from the hole? (b) Which wavelength(s) will have a strong intensity?
उत्तर
Given:-
Separation between two slits,
\[d = 0 . 5 mm = 0 . 5 \times {10}^{- 3} m\]
Wavelength of the light,
\[\lambda = 400\text{ nm to 700 nm}\]
Distance of the screen from the slit,
\[D = 50 cm = 0 . 5 m\]
Position of hole on the screen,
\[y_n = 1 mm = 1 \times {10}^{- 3} m\]
(a) The wavelength(s) will be absent in the light coming from the hole, which will form a dark fringe at the position of hole.
\[y_n = \frac{\left( 2n + 1 \right) \lambda_n}{2}\frac{D}{d}\text{, where n = 0, 1, 2, ......}\]
\[\Rightarrow \lambda_n = \frac{2}{\left( 2n + 1 \right)} \frac{y_n d}{D}\]
\[= \frac{2}{\left( 2n + 1 \right)} \times \frac{{10}^{- 3} \times 0 . 05 \times {10}^{- 3}}{0 . 5}\]
\[= \frac{2}{\left( 2n + 1 \right)} \times {10}^{- 6} m\]
\[= \frac{2}{\left( 2n + 1 \right)} \times {10}^3 nm\]
For n = 1,
\[ \lambda_1 = \left( \frac{2}{3} \right) \times 1000 = 667 nm\]
For n = 2,
\[ \lambda_2 = \left( \frac{2}{5} \right) \times 1000 = 400 nm\]
Thus, the light waves of wavelength 400 nm and 667 nm will be absent from the light coming from the hole.
(b) The wavelength(s) will have a strong intensity, which will form a bright fringe at the position of the hole.
So, \[y_n = n \lambda_n \frac{D}{d}\]
\[ \Rightarrow \lambda_n = y_n \frac{d}{nD}\]
For n = 1,
\[ \lambda_1 = y_n \frac{d}{D}\]
\[ = {10}^{- 3} \times \left( 0 . 5 \right) \times \frac{{10}^{- 3}}{0 . 5}\]
\[ = {10}^{- 6} m = 1000 nm.\]
But 1000 nm does not fall in the range 400 nm to 700 nm.
Again, for n = 2,
\[ \lambda_2 = y_n \frac{d}{2D} = 500 nm\]
So, the light of wavelength 500 nm will have strong intensity.
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