Geometrical Optics

Given, Focal length of objective lens, $f_o=10 \mathrm{~cm}$

Focal length of eye lens, $f_e=10 \mathrm{~mm}=1 \mathrm{~cm}$

Total length $=f_o+f_e=10+1=11 \mathrm{~cm}$

This is the case when the final image is formed at infinity, therefore

Angular magnification, $|m|=\left|\frac{f_o}{f_e}\right|=\frac{10}{1}=10$

From figure,

Distance of object, $u=-20 \mathrm{~cm}$

Magnification, $m=-0.5$

Magnification of lens in terms of $u$ and $f$ is given by

$\begin{aligned} m & =\frac{f}{u+f} \\ -0.5 & =\frac{f}{-20+f} \\ \Rightarrow \quad 10-0.5 f & =f \Rightarrow 1.5 f=10 \\ f & =\frac{10}{1.5} \\ f & =6.66 \mathrm{~cm} \end{aligned}$

Convex mirror always form virtual image becomes ray diverge after reflection.

According to question,

Here, $\mu=$ refractive index of lens

$\mu^{\prime}=$ refractive index of liquid focal length of lens in air $=f$

We have lens maker's formula

$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

When lens is dipped in water,

$\frac{1}{f^{\prime}}=\left(\frac{\mu_g}{\mu_l}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

or $\left(\frac{\mu_g-\mu_l}{\mu_l}\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\therefore \quad \frac{f^{\prime}}{f}=\frac{(\mu-1) \mu^{\prime}}{\left(\mu-\mu^{\prime}\right)}=\frac{-(\mu-1) \mu^{\prime}}{\left(\mu^{\prime}-\mu\right)} \quad\left(\because \mu_l=\mu^{\prime}\right)$

or $\quad f^{\prime}=\frac{-f \mu^{\prime}(\mu-1)}{\left(\mu^{\prime}-\mu\right)}$

$\begin{aligned} & \text { Given, } y=x^2 \\\\ & \qquad \frac{d y}{d x}=2 x=2 \times \frac{1}{2}=1 \\\\ & \Rightarrow \tan \theta=1=m \\\\ & \Rightarrow \quad \theta=45^{\circ} \end{aligned}$

The slope of normal at $B$

$\begin{aligned} m_2 & =-\frac{1}{m_1}=-\frac{1}{1}=-1 \\\\ \Rightarrow \quad \theta & =-45^{\circ} \end{aligned}$

From the figure angle of incidence is $45^{\circ}$. Hence, angle of reflection is $45^{\circ}$

Given, angle of prism $P_1=4^{\circ}$

$\begin{aligned} P_2 & =? \\ \eta_1 & =1.54 \\ \eta_2 & =1.72 \end{aligned}$

Angle of deviation $\delta=(n-1) A$

Dispersion produced by both the prism will be equal

$\begin{aligned} \text { i.e. } & \left(n_1-1\right) A_1=\left(n_2-1\right) A_2 \\ \Rightarrow \quad A_2 & =\frac{\left(n_1-1\right) A_1}{\left(n_2-1\right)}=\frac{(1 \cdot 54-1) \times 4}{(1 \cdot 72-1)} \\ & =3^{\circ} \end{aligned}$

Hence, the angle of prism $P_2$ is $=3^{\circ}$

For telescope

It is a case of normal adjustment.

Hence, $M=\frac{f_o}{f_e}$ also $M=\beta / \alpha$

Therefore, $\frac{\beta}{\alpha}=\frac{f_o}{f_e}$

Here $f_o=60 \mathrm{~cm}, f_e=5 \mathrm{~cm}$

$\begin{aligned} & \alpha=2^{\circ} \\ \text { Hence, } \quad & \beta=24^{\circ} \end{aligned}$

In search lights, an intense parallel beam of light is produced. In case of parabolic mirror, when source is at the focus beam of light produced over the entire cross-section is a parallel beam. But in concave spherical mirror due to its large aperture marginal rays gives a divergent beam.

The relation between angle of deviation $\delta$ for a thin prism an angle of prism and refractive index $(\mu)$ of material of prism is given by $\delta=(\mu-1) A$