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White Coherent Light (400 Nm-700 Nm) is Sent Through the Slits of a Young'S Double Slit Experiment (See the Following Figure). - Physics

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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?

योग
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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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अध्याय 17: Light Waves - Exercise [पृष्ठ ३८२]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 17 Light Waves
Exercise | Q 25 | पृष्ठ ३८२

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