A monochromatic ray of light travels through | vedclass

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

Question

(A) The number of waves $N$ in a medium of thickness $t$ is given by $N = \frac{t}{\lambda_m}$, where $\lambda_m = \frac{\lambda_0}{\mu}$ is the wavel

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(A) The number of waves $N$ in a medium of thickness $t$ is given by $N = \frac{t}{\lambda_m}$, where $\lambda_m = \frac{\lambda_0}{\mu}$ is the wavelength in the medium and $\lambda_0$ is the wavelength in vacuum.Thus, $N = \frac{t \cdot \mu}{\lambda_0}$.Given that the number of waves in the glass slab $(t_g = 4 \,cm, \mu_g = 5/3)$ is equal to the number of waves in the water column $(t_w = 5 \,cm, \mu_w = ?)$, we have:$\frac{t_g \cdot \mu_g}{\lambda_0} = \frac{t_w \cdot \mu_w}{\lambda_0}$$t_g \cdot \mu_g = t_w \cdot \mu_w$Substituting the given values:$4 	imes \frac{5}{3} = 5 	imes \mu_w$$\frac{20}{3} = 5 	imes \mu_w$$\mu_w = \frac{20}{3 	imes 5} = \frac{4}{3}$.

$A$ monochromatic ray of light travels through a glass slab and a water column. The number of waves in a glass slab of thickness $4 \,cm$ is the same as in a water column of height $5 \,cm$. If the refractive index of glass is $\frac{5}{3}$, then the refractive index of water is:

Similar Questions

$A$ microscope is focused on an object at the bottom of a bucket. If liquid with refractive index $\mu = \frac{5}{3}$ is poured inside the bucket,then the microscope has to be raised by $30\,cm$ to focus the object again. The height of the liquid in the bucket is $..........\,cm$.

The image of point $P$ when viewed from the top of the slabs will be:

Light travels through a glass plate of thickness $t$ and having refractive index $\mu$. If $c$ is the velocity of light in vacuum,the time taken by the light to travel through this thickness of glass is:

The figure shows a transparent slab of length $1 \, m$ placed in air,whose refractive index in the $x$-direction varies as $\mu = 1 + x^2$ for $0 \leq x \leq 1$. The optical path length of ray $R$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module