Question 214: Wave Optics

Question

In a Young's double slit experiment, an angular width of the fringe is 0.35$$^\circ$$ on a screen placed at 2 m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index 7/5, is $${1 \over \alpha }$$. The value of $$\alpha$$ is ___________.

Accepted Answer

Angular fringe width $$	heta  = {\lambda  \over D}$$
So $${{{	heta _1}} \over {{\lambda _1}}} = {{{	heta _2}} \over {{\lambda _2}}}$$
$${	heta _2} = {{0.35^\circ } \over {450\,nm}} 	imes {{450\,nm} \over {7/5}} = 0.25^\circ  = {1 \over 4}$$
So $$\alpha  = 4$$

Wave Optics

Integer / Numeric Answer Type