Question

In YDSE, when dichromatic source of light is used we get interference pattern which is obtained by combination of interference pattern due to each wavelength. A black line is obtained on screen where a minima due to each wavelength coincides. Similarly, when polychromatic source is taken say white light source, resulting interference pattern is superimposition of interference pattern from each wavelength as a result at which we get white central bright fringe surrounded by few coloured fringes on both sides of central bright fringe and then uniform illumination of screen. In a typical YDSE set up, distance between slits(D) is 1mm and screen is placed at a distance (D) of 1 m from slits. The source of light used is dichromatic emitting waves of wavelength ${\lambda }_1$ = 500 nm and ${\lambda }_2$ = 700 nm.

The position of first black line from central maxima is

• A. $28\times {10}^{-3}m$
• B. $17.5\times {10}^{-4}m$
• C. $28\times {10}^{-5}m$
• D. None

Answer 1

The correct option is B $17.5\times {10}^{-4}m$

Dark fringe position $y=\frac{(2n-1)D\lambda }{2d}$

Let $n_{1}and{n}_2$ coincide $(2n_1-1){\lambda }_1=(2n_2-1){\lambda }_2$

$\frac{(2n_1-1)}{(2n_2-1)}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{7}{5}$

$\Rightarrow n_1=4,n_2=3forfirstblackline$

4th dark of ${\lambda }_1=\frac{(2\times 4-1)D{\lambda }_1}{2d}=\frac{7\times 1\times 500\times {10}^{-9}}{2\times {10}^{-3}}=17.5\times {10}^{-4}m$