NEET 2020 Physics Question 165 Solution

In Young's double-slit experiment, the fringe width $$\beta$$ (distance between consecutive bright or dark fringes) is given by: $$\beta = \frac{\lambda D}{d}$$, where $$\lambda$$ is the wavelength of light, $$D$$ is the distance between slits and screen, and $$d$$ is the slit separation.

- Separation between coherent sources is halved: $$d' = \frac{d}{2}$$ - Distance of the screen is doubled: $$D' = 2D$$ - Wavelength $$\lambda$$ remains unchanged.

$$\beta' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{\frac{d}{2}} = \frac{2}{\frac{1}{2}} \left(\frac{\lambda D}{d}\right) = 4 \left(\frac{\lambda D}{d}\right) = 4\beta$$.

Thus, the fringe width becomes four times its original value, matching Option 3.