A beaker containing liquid is placed on a tab | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The relation between real depth $(h)$ and apparent depth $(h')$ is given by $h = \mu h'$,where $\mu$ is the refractive index.In the first case,the

Vedclass · Accepted Answer

$\frac{d - b}{d - c - b + a}$

Vedclass · Answer

(A) The relation between real depth $(h)$ and apparent depth $(h')$ is given by $h = \mu h'$,where $\mu$ is the refractive index.In the first case,the real depth of the liquid is $h_1 = \mu(b - a)$,where $(b - a)$ is the apparent depth.In the second case,the real depth of the liquid is $h_2 = \mu(d - c)$,where $(d - c)$ is the apparent depth.The difference in real depth is $\Delta h = h_2 - h_1 = \mu(d - c - b + a)$.This difference in real depth is equal to the thickness of the additional liquid added,which is $\Delta h = d - b$.Equating the two expressions: $\mu(d - c - b + a) = d - b$.Therefore,the refractive index is $\mu = \frac{d - b}{d - c - b + a}$.

Similar Questions

$A$ light beam is incident on a denser medium whose refractive index is $1.414$ at an angle of incidence $45^o$. Find the ratio of the width of the refracted beam in the medium to the width of the incident beam in air.

When light is refracted from air into glass,

According to the Cartesian sign convention in ray optics,

The apparent depth of water in a cylindrical water tank of diameter $2R \, cm$ is reducing at the rate of $x \, cm/minute$ when water is being drained out at a constant rate. The amount of water drained in $c.c./minute$ is ($n_1 =$ refractive index of air,$n_2 =$ refractive index of water).

Vedclass Test Series

Exam Paper Generator

Online Exam Module