In the figure below,$PQRS$ denotes the path followed by a ray of light as it travels through three media in succession. The absolute refractive indices of the media are $\mu_1, \mu_2$ and $\mu_3$,respectively. (The line segment $RS$ in the figure is parallel to $PQ$). Then,

$A$ 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.

The incorrect statement about the refractive index for a pair of media is:

$A$ monochromatic light wave with wavelength $\lambda_1$ and frequency $v_1$ in air enters another medium. If the angle of incidence and angle of refraction at the interface are $45^{\circ}$ and $30^{\circ}$ respectively,then the wavelength $\lambda_2$ and frequency $v_2$ of the refracted wave are :

If ${ }_{i} \mu_{j}$ represents the refractive index when a ray goes from medium $i$ to medium $j$,then the product ${ }_2 \mu_1 \times { }_3 \mu_2 \times { }_4 \mu_3$ is equal to:

Two light beams fall on a transparent material block at points $1$ and $2$ with angles $\theta_1$ and $\theta_2$ respectively,as shown in the figure. After refraction,the beams intersect at point $3$,which is exactly on the interface at the other end of the block. Given: the distance between $1$ and $2$ is $d = 4\sqrt{3} \text{ cm}$ and $\theta_1 = \theta_2 = \cos^{-1}\left(\frac{n_2}{2n_1}\right)$,where $n_2$ is the refractive index of the block and $n_1$ is the refractive index of the outside medium $(n_2 > n_1)$. Find the thickness of the block in $\text{cm}$.