This solution calculates the time a light ray spends inside a glass slab.
Apply Snell's Law:
The relationship between the angle of incidence ($i$) in air and the angle of refraction ($r$) inside the glass slab ($\mu$) is given by:
$ \sin i = \mu \sin r $
Solving for $\sin r$:
$ \sin r = \frac{\sin i}{\mu} $
Calculate the distance travelled inside the slab:
Let the path length of the light ray inside the slab be $L$. The thickness of the slab is $t$. The angle the light ray makes with the normal inside the slab is $r$.
Consider the right-angled triangle formed by the thickness $t$ (adjacent to angle $r$) and the path length $L$ (hypotenuse).
$ \cos r = \frac{t}{L} $
Therefore, the distance $L$ is:
$ L = \frac{t}{\cos r} $
We need $\cos r$. Using the identity $\sin^2 r + \cos^2 r = 1$:
$ \cos r = \sqrt{1 - \sin^2 r} $
Substitute $\sin r = \frac{\sin i}{\mu}$:
$ \cos r = \sqrt{1 - \left(\frac{\sin i}{\mu}\right)^2} = \sqrt{\frac{\mu^2 - \sin^2 i}{\mu^2}} = \frac{\sqrt{\mu^2 - \sin^2 i}}{\mu} $
Now substitute $\cos r$ back into the equation for $L$:
$ L = \frac{t}{\frac{\sqrt{\mu^2 - \sin^2 i}}{\mu}} = \frac{\mu t}{\sqrt{\mu^2 - \sin^2 i}} $
Calculate the speed of light inside the slab:
The speed of light ($v$) in the glass slab is:
$ v = \frac{c}{\mu} $
Calculate the total time spent:
The time ($T$) is the distance $L$ divided by the speed $v$:
$ T = \frac{L}{v} = \frac{\left(\frac{\mu t}{\sqrt{\mu^2 - \sin^2 i}}\right)}{\left(\frac{c}{\mu}\right)} $
$ T = \frac{\mu t}{\sqrt{\mu^2 - \sin^2 i}} \times \frac{\mu}{c} $
$ T = \frac{\mu^2 t}{c\sqrt{\mu^2 - \sin^2 i}} $
The total time spent by the ray inside the slab is $\frac{\mu^2 t}{c\sqrt{\mu^2 - \sin^2 i}}$.
Two points of monochromatic and coherent sources of light of wavelength $\lambda$ each, are placed as shown in figure. The initial phase difference between the sources is zero, ($D \gg d$). Mark the correct statement(s).