Question 16: Wave Optics

Question

In YDSE, monochromatic light falls on a screen 1.80 m from two slits separated by 2.08 mm . The first and second order bright fringes are separated by 0.553 mm . The wavelength of light used is

Accepted Answer

In the Young's Double Slit Experiment (YDSE), for monochromatic light shining on a screen 1.80 m away from two slits separated by 2.08 mm, the separation of the first and second order bright fringes is 0.553 mm. To find the wavelength of light used, we use the following calculations:
The position of the first bright fringe is given by:
$ y_1 = \frac{\lambda D}{d} $
The position of the second bright fringe is:
$ y_2 = \frac{2\lambda D}{d} $
The separation between the first and second order bright fringes is:
$ y_2 - y_1 = \frac{2\lambda D}{d} - \frac{\lambda D}{d} = \frac{\lambda D}{d} $
From this, we can derive the formula for the wavelength $\lambda$ as:
$ \lambda = \frac{(y_2 - y_1) d}{D} $
Substituting the given values into this equation:
$ \lambda = \frac{0.553 	imes 10^{-3} 	imes 2.08 	imes 10^{-3}}{1.8} $
Solving this gives:
$ \lambda = 639 	imes 10^{-9} 	ext{ m} $
Therefore, the wavelength of the light used is:
$ \lambda = 639 	ext{ nm} $

Answer

520 nm

Answer

639 nm

Answer

715 nm

Answer

None of these

Wave Optics

In YDSE, monochromatic light falls on a screen 1.80 m from two slits separated by 2.08 mm . The first and second order bright fringes are separated by 0.553 mm . The wavelength of light used is

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