A beam of light is incident vertically on a glass slab of thickness $1 \mathrm{cm}$, and refractive index $1.5$. A fraction 'A' is reflected from the front surface while another fraction 'B' enters the slab and emerges after reflection from the back surface. Time delay between them is

When light passes through a medium with α different refractive index, part of the light is reflected, and part is transmitted (refracted). In this case, a beam of light is incident vertically on a glass slab of thickness 1 cm and refractive index 1.5. A fraction ‘A’ of the light is reflected from the front surface, while another fraction ‘B’ enters the slab, reflects from the back surface, and emerges.
The time delay between the reflected beam (‘A’) and the transmitted and reflected beam (‘B’) is due to the  extra distance travelled by the transmitted beam inside the glass slab.
Let’s consider the path of the transmitted and reflected beam (‘B’): 
1. The beam enters the front surface of the slab and travels through the glass for a distance equal to the thickness of the slab, which is 1 cm.
2. The beam reflects from the back surface of the slab and starts travelling back through the glass again.
3. After travelling back through the glass slab, the beam exits the front surface and emerges.
Since the refractive index of the glass slab is 1.5, the speed of light inside the glass (v_g) is:
${\mathrm{v}}_{\mathrm{g}}=\frac{\mathrm{c}}{\mathrm{\mu }}$
where,
c is the speed of light in vacuum,
μ is the refractive index of the glass (1.5).
${\mathrm{v}}_{\mathrm{g}}=\frac{3\times {10}^8}{1.5}=2\times {10}^8\mathrm{m}/\mathrm{s}$
Now, let’s calculate the time taken for the transmitted and refl ected beam (‘B’) inside the glass slab:
Time delay (t) = Distance travelled inside the glass / Speed of light inside the glass
$\begin{array}{c}\mathrm{t}=\frac{\mathrm{s}}{{\mathrm{v}}_{\mathrm{g}}}=\frac{\left(2\times {10}^{-2}\right)}{2\times {10}^8}\\ ={10}^{-10}\mathrm{s}\end{array}$