In Young's double-slit experiment, the distance between slits and the screen is 1 m and monochromatic light of wavelength 600 nm is being used. A person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied, the interference pattern disappears for a particular distance $$d_0$$ between the slits. If the angular resolution of the eye is $$\frac{1}{60}°$$, then the value of $$d_0$$ is close to

We have monochromatic light of wavelength $$\lambda = 600 \text{ nm}=600\times10^{-9}\,\text{m}$$ falling on a pair of slits separated by a distance $$d$$. The interference pattern is observed on a screen placed at a distance $$L=1\ \text{m}$$, and a person keeps his eye very close to the slits to look at that pattern.

For a double-slit arrangement, the angular position of the $$m^{\text{th}}$$ bright fringe is given, for small angles, by the well-known formula

$$\theta_m \approx \frac{m\lambda}{d}\,.$$

Hence the angular separation between two successive bright fringes is obtained by writing

$$\Delta\theta=\theta_{m+1}-\theta_m =\frac{(m+1)\lambda}{d}-\frac{m\lambda}{d} =\frac{\lambda}{d}\,.$$

The eye can resolve two distinct bright lines only if the angle between them is not smaller than the eye’s minimum resolvable angle. The question tells us that the angular resolution of the eye is

$$\theta_{\min}=\frac{1}{60}^{\circ}\,.$$

First we change this to radians because our interference formula is in radians. We use the relation $$1^{\circ}=\frac{\pi}{180}\ \text{rad}$$, so

$$\theta_{\min}= \frac{1}{60}\times\frac{\pi}{180} =\frac{\pi}{10800}\ \text{rad}\,.$$ Numerically,

$$\theta_{\min}= \frac{3.1416}{10800}\approx2.91\times10^{-4}\ \text{rad}\,.$$

As the slit separation $$d$$ is gradually increased, the angular fringe spacing $$\Delta\theta=\frac{\lambda}{d}$$ steadily decreases. The fringes will cease to be distinguishable exactly when the separation between adjacent bright fringes becomes equal to the eye’s limiting angle, that is when

$$\Delta\theta=\theta_{\min}\; \Longrightarrow\; \frac{\lambda}{d_0}= \theta_{\min}\,.$$ Therefore

$$d_0=\frac{\lambda}{\theta_{\min}} =\frac{600\times10^{-9}} {\dfrac{\pi}{10800}} =600\times10^{-9}\times\frac{10800}{\pi}\,.$$ Carrying out the multiplications,

$$600\times10800=6\,480\,000, \quad 6\,480\,000\times10^{-9}=6.48\times10^{-3}\,,$$ so

$$d_0=\frac{6.48\times10^{-3}}{3.1416} \approx2.06\times10^{-3}\,\text{m} =2.06\ \text{mm}\,.$$

This value is closest to $$2\ \text{mm}$$ among the given options.

Hence, the correct answer is Option C.