Ray Optics

Given, focal length of convex lens,

$ f=25 \mathrm{~cm} $

Refractive index, $\mu_g=1.5$

By using lens Maker's formula,

$ \frac{1}{f}=\left(\mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

When cover lens is immersed in water, then focal length $\left(f_w\right)$ is given as

$ \frac{1}{f_w}=\mu\left(\frac{\mu_g}{\mu_w}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Dividing Eq. (i) by Eq. (ii), we have

$ \begin{aligned} & \frac{\frac{1}{f}}{\frac{1}{f_w}}=\frac{\left(\mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)}{\left(\frac{\mu_g}{\mu_w}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)} \\ \Rightarrow \quad & \frac{f_w}{f}=\frac{\left(\mu_g-1\right) \mu_w}{\left(\mu_g-\mu_w\right)}=\frac{(1.5-1) \frac{4}{3}}{\left(1.5-\frac{4}{3}\right)}=4 \\ \Rightarrow \quad & f_w=4 f=4 \times 25=100 \mathrm{~cm} \end{aligned} $

∴ Change in focal length

$ =f_w-f=100-25=75 \mathrm{~cm} $

Given, focal length of convex lens,

$ f=\frac{20}{3} \mathrm{~cm} $

Radii of curvature of double convex lens,

$ R_1=5 \mathrm{~cm}, R_2=-10 \mathrm{~cm} $

By lens Maker's formula,

$ \begin{aligned} & & \frac{1}{f} & =(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \Rightarrow & & \frac{1}{20} & =(\mu-1)\left(\frac{1}{5}-\frac{1}{-10}\right) \\ \Rightarrow & & \frac{3}{20} & =(\mu-1)\left(\frac{1}{5}+\frac{1}{10}\right) \\ \Rightarrow & & \frac{3}{20} & =(\mu-1)\left(\frac{3}{10}\right) \\ \Rightarrow & & \frac{1}{2} & =\mu-1 \\ \Rightarrow & & \mu & =1+\frac{1}{2} \\ & & & =\frac{3}{2}=1.5 \end{aligned} $

$ \begin{aligned} &\text { Use the formula, }\\ &\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R} \end{aligned} $

Given, $n_1=2, n_2=\sqrt{3}$

If $i_c$ be the critical angle, then

$ \begin{array}{r} \quad \sin i_c=\frac{1}{2} n_1=\frac{1}{\frac{n_1}{n_2}}=\frac{n_2}{n_1}=\frac{\sqrt{3}}{2} \\ \Rightarrow \quad \sin i_c=\sin 60^{\circ} \Rightarrow i_c=60^{\circ} \end{array} $

By using lens formula,

$ \begin{aligned} \frac{1}{f} & =\frac{1}{v}-\frac{1}{u} \\ & =\frac{1}{20}-\frac{1}{(-5)} \quad\binom{\because v=20 \mathrm{~cm}}{u=-5 \mathrm{~cm}} \\ & =\frac{1+4}{20}=\frac{1}{4} \end{aligned} $

When the glass plate of thickness $t(1.4 \mathrm{~cm})$ is inserted between the lens and the screen, then the shift in position of image

$ \begin{aligned} & =t\left(1-\frac{1}{\mu}\right) \\ & =1.4\left(1-\frac{1}{7 / 5}\right) \quad\left(\because \mu=\frac{7}{5}\right) \\ & =1.4\left(\frac{2}{7}\right)=0.4 \mathrm{~cm} \end{aligned} $

Thus, the lens forms the image at a distance of $20-0.4=19.6 \mathrm{~cm}$ from itself.

Hence, using again lens formula,

$ \begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow & \frac{1}{u}=\frac{1}{v}-\frac{1}{f}=\frac{1}{19.6}-\frac{1}{4}=\frac{3.9}{19.6}=\frac{39}{196} \\ \Rightarrow & u=\frac{196}{39}=5.02 \mathrm{~cm} \end{aligned} $

Refractive index, $\mu=1.5$

Since, one side of given lens is flat and other side is convex with a radius of curvature $R$.

$ \therefore R_1=R \text { and } R_2=\infty $

Position of object, $u=-60 \mathrm{~m}$

Position of image, $v=120 \mathrm{~cm}$

$ \begin{aligned} &\text { Using lens Maker's formula, we get }\\ &\begin{array}{ll} & \frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \Rightarrow & \frac{1}{v}-\frac{1}{u}=(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right) \\ & \quad\left(\because \frac{1}{f}=\frac{1}{v}-\frac{1}{u}\right) \\ \Rightarrow & \frac{1}{120}-\frac{1}{-60}=(1.5-1) \cdot \frac{1}{R} \\ \Rightarrow & \frac{3}{120}=\frac{0.5}{R} \Rightarrow \frac{1}{40}=\frac{0.5}{R} \\ \Rightarrow & R=40 \times 0.5=20 \mathrm{~cm} \end{array} \end{aligned} $

For water, real depth $=12 \mathrm{~cm}$

Apparent depth $=9 \mathrm{~cm}$

Since, refractive index, $\mu_w=\frac{\text { Real depth }}{\text { Apparent depth }}$

$\Rightarrow \quad \mu_{\mathrm{w}}=\frac{12}{9}$

when water is replaced by liquid of refractive index $\mu=1.5$, we have

New apparent depth $=\frac{\text { real depth }}{\mu_l}=\frac{12}{1.5}=8 \mathrm{~cm}$

Hence, required shifted distance $=9-8=1 \mathrm{~cm}$

For double convex lens,

$R_1=4 \mathrm{~cm}, R_2=-8 \mathrm{~cm}$

Refractive index, $\mu=1.5$

By lens maker's formula, focal length is given by

$\begin{aligned} \frac{1}{f} & =(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ & =(1.5-1)\left(\frac{1}{4}-\frac{1}{-8}\right)=0.5\left(\frac{1}{4}+\frac{1}{8}\right) \\ & =0.5\left(\frac{3}{8}\right)=\frac{3}{16} \\ \Rightarrow \quad f & =\frac{16}{3} \mathrm{~cm} \\ & =5.33 \mathrm{~cm} \end{aligned}$

From the given situation abṣolute refractive index of denser medium, ${ }^a \mu_d=2$

Absolute refractive index of rarer medium, ${ }^a \mu_r=1$

$\therefore$ Refractive index of denser medium with respect to rarer medium

${ }^r \mu_d=\frac{{ }^a \mu_d}{{ }^a \mu_r}=\frac{2}{1}=2 \Rightarrow{ }^r \mu_d=2$

If $i_c$ be the critical angle, then

$\begin{aligned} \sin i_c & =\frac{1}{{ }_r^r \mu_d}=\frac{1}{2}=\sin 30^{\circ} \\ \Rightarrow \quad i_c & =30^{\circ} \end{aligned}$

By using lens Maker formula,

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

where,

$\begin{aligned} & f=\text { focal length, } \\ & \mu=\text { refractive index, } \end{aligned}$

$R_1$ and $R_2=$ radii of curvature.

If light changes from red to blue, refractive index of light also changes. Hence, focal length will also change.

Hence, Assertion and Reason both are false.

Given, wavelength of light $A, \lambda_A=300 \mathrm{~nm}$

$=300 \times 10^{-9} \mathrm{~m}$

Frequency of light in $A, f_A=5 \times 10^{14} \mathrm{~Hz}$

Ratio of speed of light $\frac{v_A}{v_B}=\frac{4}{5}$

Let $\mu_A, \mu_{\text {air }}$ and $\mu_B$ are refractive indexes of medium $A$, air and $B$ respectively.

$\begin{aligned} & \because \text { Absolute refractive index }=\frac{\mu_A}{\mu_{\text {air }}} \\ & =\frac{\text { Speed of light in air }(c)}{\text { Speed of light in medium } A\left(v_A\right)} \\ & \Rightarrow \quad \mu_A=\frac{3 \times 10^8}{f_A \lambda_A} \\ & \Rightarrow \quad \mu_A=\frac{3 \times 10^8}{5 \times 10^{14} \times 300 \times 10^{-9}}=0.2 \times 10^1=2 \\ \end{aligned}$

Now, $\frac{\mu_B}{\mu_A}=\frac{v_A}{v_B}$

$\Rightarrow \quad \mu_B=\mu_A\left(\frac{v_A}{v_B}\right)=2\left(\frac{4}{5}\right)=\frac{8}{5}=1.6 $

Let $\delta_m$ be the angle of minimum deviation.

As we know that,

$ \mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}} $

Given, $\quad \mu=\sqrt{2}$ and $A=\delta_m$

$ \begin{array}{ll} \Rightarrow & \sqrt{2}=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \frac{A}{2}}=\frac{\sin A}{\sin \frac{A}{2}} \\ \Rightarrow & \sqrt{2}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} \\ & \quad \because \sin 2 A=2 \sin A \cos A] \\ & \quad \sqrt{2}=2 \cos \frac{A}{2} \Rightarrow \cos \frac{A}{2}=\frac{1}{\sqrt{2}} \\ \Rightarrow & \quad \frac{A}{2}=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) \Rightarrow \frac{A}{2}=45^{\circ} \\ \Rightarrow & A=90^{\circ} \end{array} $

For dispersion without deviation through a prism combination,

$ \begin{aligned} \delta_{m_1} & =\delta_{m_2} \Rightarrow A_1\left(\mu_1-1\right)=A_2\left(\mu_2-1\right) \\ \Rightarrow A_2 & =A_1 \frac{\left(\mu_1-1\right)}{\left(\mu_2-1\right)} \end{aligned} $

Here, $A_1=9^{\circ}, \mu_1=1.4$ and $\mu_2=1.6$

$ \begin{aligned} \therefore \quad A & =A_2=9 \times \frac{(1.4-1)}{(1.6-1)} \\ & =9 \times \frac{0.4}{0.6}=6^{\circ} \end{aligned} $

The given situation is shown in the following figure.

$A B \rightarrow$ object, $A^{\prime} B^{\prime} \rightarrow$ image

$ A B=l, A^{\prime} B^{\prime}=l^{\prime}, u=-X, $

Hence, by mirror formula,

$ \begin{align*} & \frac{1}{-F} =\frac{1}{v}+\frac{1}{(-X)} \quad\left[\because \frac{1}{f}=\frac{1}{v}+\frac{1}{u}\right] \\ \Rightarrow \frac{1}{v} & =\frac{1}{x}-\frac{1}{F} \Rightarrow \frac{1}{v}=\frac{F-X}{F X} \\ \Rightarrow & v =\frac{F X}{F-x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{align*} $

Longitudinal magnification for small object is given as

$ \begin{aligned} M & =\frac{l^{\prime}}{l}=-\frac{v^2}{u^2}=-\frac{\left(\frac{F X}{F-X}\right)^2}{(-X)^2} \\ & =-\frac{f^2 X^2}{(F-X)^2 X^2}=-\left(\frac{F}{F-X}\right)^2 \end{aligned} $

Magnitude of longitudinal magnification, $\frac{l^{\prime}}{l}=\left(\frac{F}{F-X}\right)^2$

The given situation is shown in the following figure.

For the ray incident at face $A C$ to be total internally reflected, if Angle of incidence $=$ Critical angle

$ \begin{aligned} & \Rightarrow \quad i=i_c \Rightarrow \sin (i)=\sin \left(i_c\right) \\ & \Rightarrow \quad \sin (90-\theta)=\frac{\mu_{\text {rarer }}}{\mu_{\text {denser }}} \\ & \Rightarrow \quad \cos (\theta)=\frac{\mu_l}{\mu_g} \Rightarrow \cos (\theta)=\frac{1.2}{1.5} \\ & \Rightarrow \quad \cos \theta=\frac{4}{5} \Rightarrow \cos \theta=0.8 \\ & \Rightarrow \quad \theta=\cos ^{-1}(0.8) \end{aligned} $

Magnification of a telescope for normal setting is

$ m=\frac{f_0}{f_e}=\frac{100}{5}=20 $

The given situation is shown below

For convex lens, $u=-8 \mathrm{~cm}, f=4 \mathrm{~cm}$

$ \begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{1}{4}+\frac{1}{-8} \Rightarrow \frac{1}{v}=\frac{1}{8} \\ & \Rightarrow v=8 \mathrm{~cm} \end{aligned} $

For concave lens, image of convex lens acts like a virtual object.

$ \begin{aligned}So,\,\, & u=+2 \mathrm{~cm}, f=-4 \mathrm{~cm} \\ & \frac{1}{v}=\frac{1}{-4}+\frac{1}{2}=\frac{1}{4} \mathrm{~cm} \Rightarrow v=4 \mathrm{~cm} \end{aligned} $

So, distance from final image and object

$ =(8+6+4) \mathrm{cm}=18 \mathrm{~cm} $

The given situation is shown in the following figure

Given, $\quad v=+f, f=+f$

By $\quad \frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

We have, $\frac{1}{f}-\frac{1}{f}=\frac{1}{u} \Rightarrow \frac{1}{u}=0$

$ \text { ⇒ } \quad u=\infty $

So, object is at $\infty$ distance on right side of lens.