Geometrical Optics

To replace a bi-convex lens with another convex lens that has different radii of curvature on each side (i.e., $ R_1 \neq R_2 $), while maintaining the same lens power, the radii must satisfy a specific relationship.

For the current bi-convex lens, both radii of curvature are given as $ \frac{1}{6} \, \text{cm} $. The lens formula for power is tied to the radii through:

$ \frac{1}{f_1} = (\mu - 1) \left( \frac{2}{R} \right) $

When this lens is replaced by a lens with different radii ($ R_1 $ and $ R_2 $), the condition that has to be met is:

$ \frac{1}{f_2} = (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) $

For power equivalence, the expressions for $ \frac{1}{f_1} $ and $ \frac{1}{f_2} $ must be equal:

$ (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) = (\mu - 1) \left( \frac{2}{R} \right) $

This simplifies to:

$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{R} $

Given $ R = \frac{1}{6} \, \text{cm} $, it follows that:

$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{\left(\frac{1}{6}\right)} $

Thus, simplifying the equation gives:

$ \frac{1}{R_1} + \frac{1}{R_2} = 12 $

Therefore, any valid pair of radii $ R_1 $ and $ R_2 $ must satisfy this equation.

To determine the image distance when a spherical surface separates two media with refractive indices of 1 and 1.5:

Use the lens maker's formula for a spherical surface:

$ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} $

Where:

$\mu_1$ = 1 (refractive index of the first medium)

$\mu_2$ = 1.5 (refractive index of the second medium)

$u$ = -0.2 m (object distance, negative as per convention)

$R$ = 0.4 m (radius of curvature)

Substituting the known values:

$ \frac{1.5}{v} - \frac{1}{-0.2} = \frac{1.5 - 1}{0.4} $

Simplify the equation:

$ \frac{1.5}{v} = \frac{0.5}{0.4} - \frac{1}{0.2} $

Calculate:

$ \frac{1.5}{v} = -\frac{1.5}{0.4} $

Solve for $v$:

$ v = -0.4 \, \text{m} $

The negative sign indicates that the image is 0.4 meters to the left of the spherical surface.

Location of image of A :-

$\begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}-\frac{1}{-30}=\frac{1}{20} \Rightarrow \frac{1}{v}=\frac{1}{60} \Rightarrow v=60 \mathrm{~cm} \\ & \therefore m=2 \end{aligned}$

Since size of object is small wrt the location hence

$\begin{aligned} & d v=m^2 d u \Rightarrow d v=4 \times 1=4 \mathrm{~cm} \\ & h_i=m h_0 \Rightarrow h_i(d y)=2 \times 2=4 \mathrm{~cm} \end{aligned}$

$\therefore$ Angle made with principle axis $=-45^{\circ}$

Using lens formula,

For Ist case : ${1 \over f} = \left( {{a^{{\mu _g}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

where, ${R_1},{R_2}$ = radii of refracting surfaces

$f$ = focal length

${a^{{\mu _g}}} = {{^{{\mu _g}}} \over {^{{\mu _a}}}} = {{1.5} \over 1} = 1.5$

So, ${1 \over {24}} = (1.5 - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$ (As f = 24 cm given)

$ \Rightarrow {1 \over {24}} = 0.5\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

$ \Rightarrow {1 \over {12}} = \left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

$ \Rightarrow {1 \over {{R_1}}} - {1 \over {{R_2}}} = {1 \over {12}}$ .... (i)

For IInd case:

${1 \over f} = \left( {w{\mu _g} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

$ \Rightarrow {1 \over f} = \left( {{{1.5} \over {1.33}} - 1} \right)\left( {{1 \over {12}}} \right)$ ..... from (i)

$ \Rightarrow {1 \over f} = {{17} \over {133}} \times {1 \over {12}} \Rightarrow f = {{133 \times 12} \over {17}}$

$ \Rightarrow f = 93.88$ cm

$\approx$ 94 cm

Hence, option 1 is correct.

Using lens maker formula,

${{{\mu _2}} \over v} - {{{\mu _1}} \over 4} = {{{\mu _2} - {\mu _1}} \over R}$

For the image formed by the surface on the right side.

${\mu _1} = 1,\,{\mu _2} = 1.5,\,u = {{ - R} \over 2},{R_1} = - R$

so, ${{1.5} \over v} - {1 \over { - R/2}} = {{1.5 - 1} \over { - R}}$

$ \Rightarrow {{1.5} \over v} + {2 \over R} = {{ - 1} \over {2R}}$

$ \Rightarrow {{1.5} \over v} = {{ - 5} \over {2R}} \Rightarrow v = {{ - 3R} \over 5}$

Here '$-$' sign tells that image is formed left to B.

So, the distance of image from $P, = {{R - 3R} \over \Delta } = {{2R} \over 5}$

Now, for the surface on the left side,

${\mu _1} = 1,{\mu _2} = 1.5,\mu = {{3R} \over 2},{R_2} = R$

So, ${{1.5} \over v} - {1 \over {{{3R} \over 2}}} = {{1.5 - 1} \over R}$

$ \Rightarrow {{1.5} \over v} - {2 \over {3R}} = {1 \over {2R}}$

$ \Rightarrow {{1.5} \over v} = {1 \over R}\left( {{1 \over 2} + {2 \over 3}} \right)$

$ \Rightarrow {{1.5} \over v} = {7 \over {6R}}$

$ \Rightarrow v = {{9R} \over 7}$

So distance from $P = {{9R} \over 7} - R = {{2R} \over 7}$

So the distance between two images,

$ = {{2R} \over 5} - {{2R} \over 7} = 0.4R - 0.2857R$

$ \approx 0.4R - 0.286R = 0.114R$.

Here,

we will use lens maker formula,

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

So, the focal length of L$_1$ remains the same as the original lens because the curvature of the surfaces does not change

So, ${\left( f \right)_{{L_1}}} = f$ .... (1)

Now, for L$_3$, L$_3$ is plano-convex lens

${R_1} = R,\,{R_2} = \infty $

So, ${1 \over {{{\left( f \right)}_{{L_3}}}}} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over \infty }} \right)$

$ \Rightarrow {1 \over {{{\left( f \right)}_{{L_3}}}}} = \left( {\mu - 1} \right){1 \over R}$ ..... (2)

For whole double convex lens,

${1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over { - R}}} \right)$ (As ${R_1} = R$, ${R_2} = - R$)

$ \Rightarrow {1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over R} + {1 \over R}} \right)$

$ \Rightarrow {1 \over f} = \left( {\mu - 1} \right)\left( {{2 \over R}} \right)$ .... (3)

From (2) and (3),

${1 \over {{{\left( f \right)}_{{L_3}}}}} = {1 \over {2f}} \Rightarrow {\left( f \right)_{{L_3}}} = 2f$

Hence, ${{{{\left( f \right)}_{{L_1}}}} \over {{{\left( f \right)}_{{L_3}}}}} = {f \over {2f}} \quad\text{....(from (1)})\quad= 1:2$

$(u+f)(v-f)=f^2$

$\begin{aligned} & \mathrm{m}=-3=-\frac{\mathrm{v}}{\mathrm{u}} \text { and } \mathrm{v}-\mathrm{u}=20 \mathrm{~cm} \\ & \mathrm{f}=\frac{\mathrm{vu}}{\mathrm{v}+\mathrm{u}}=\frac{(-30)(-10)}{-30-10} \\ & \therefore \mathrm{R}=+15 \end{aligned}$

$\begin{aligned} & \frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}} \frac{\mathrm{n}_2-\mathrm{n}_1}{\mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}-\frac{1.3}{-13}=\frac{0.1}{\mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}=\frac{1-\mathrm{R}}{10 \mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}=\frac{1-\mathrm{R}}{10 \mathrm{R}} \\ & \mathrm{~m}=\frac{\mathrm{v} / \mathrm{n}_2}{\mathrm{u} / \mathrm{n}_1} \\ & -2 \times \frac{(-13)}{1.3}=\frac{10 \mathrm{R}}{1-\mathrm{R}} \\ & \mathrm{R}=\frac{2}{3} \mathrm{~cm} \end{aligned}$

We know, for thin prism,

$\mu = {{A + \delta } \over A}$

where, $\mu$ = refractive index

A = Angle of prism

$\delta$ = Angle of deviation

We can write, $\mu A = A + \delta $

$ \Rightarrow \delta = (\mu - 1)A$

given, ${\delta _{net}} = 0$

$ \Rightarrow ({\mu _1} - 1){A_1} - ({\mu _2} - 1){A_2} = 0$

$ \Rightarrow (1.54 - 1)4 - (1.72 - 1){A_2} = 0$

$ \Rightarrow {A_2} = {{54 \times 4} \over {72}} \Rightarrow {A_2} = 3^\circ $

from $\Delta OCB,\,\tan \theta = {{CB} \over {OC}} = {R \over R} = 1$

$ \Rightarrow \theta = 45^\circ $

We know, $\sin c = {1 \over \mu }$

for minimum $\mu$, sinc maximum $\Rightarrow$ C maximum

Here, ${c_{\max }} = \theta = 45^\circ $

So, $\mu = {1 \over {\sin 45^\circ }} = {1 \over {{1 \over {\sqrt 2 }}}}$

$ \Rightarrow \mu = \sqrt 2 $

$\begin{aligned} &\mathrm{H}=18 \mathrm{~km}\\ &\text { Size of camera film }=2 \mathrm{~cm} \times 2 \mathrm{~cm}\\ &\begin{aligned} & \mathrm{A}_{\text {image }}=400 \mathrm{~km}^2 \\ & \mathrm{x}=20 \times 10^3 \mathrm{~m}=2 \times 10^4 \mathrm{~m} \\ & \mathrm{y}=2 \times 10^{-2} \mathrm{~m} \\ & \frac{\mathrm{x}}{\mathrm{y}}=10^6=\frac{18 \mathrm{Km}}{\mathrm{f}} \\ & \mathrm{f}=18 \times 10^{-3} \mathrm{~m}=18 \mathrm{~mm} \\ & \mathrm{f}=1.8 \mathrm{~cm} \end{aligned} \end{aligned}$

$ P_1 = 2.5\,D,\quad f_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4\,m $

After an increase in optical power by 0.1 D:

$ P_2 = 2.5\,D + 0.1\,D = 2.6\,D,\quad f_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846\,m $

The change in focal length is:

$ \Delta f = f_1 - f_2 = 0.4\,m - 0.3846\,m \approx 0.0154\,m $

The relative decrease in focal length is then calculated as:

$ \text{Relative decrease} = \frac{\Delta f}{f_1} = \frac{0.0154}{0.4} \approx 0.0385 \approx 0.04 $

Thus, the relative decrease in focal length is approximately 0.04.

$ \frac{1}{f} = \frac{2(n_2 - n_1)}{n_1 R} $

For a thin plano‑convex lens made of glass (refractive index $n_2 = 1.5$) immersed in a liquid (refractive index $n_1 = 1.2$), when the plane side is silver‐coated, a ray entering the curved surface from the liquid is refracted into the glass, reflects off the coated plane (thereby reversing its direction) and finally refracts back into the liquid at the same curved surface. In the paraxial approximation the net effect is equivalent to a mirror in the liquid with an effective focal length $f$.

A ray that is incident parallel to the optical axis (at height $h$) will, after this combined refraction–reflection–refraction process, emerge with an angular deviation corresponding to a mirror having a power given by:

$ \frac{h}{f} = \frac{2(n_2 - n_1)}{n_1R}\, h\,. $

This gives the relation:

$ \frac{1}{f} = \frac{2(n_2 - n_1)}{n_1 R}\,. $

Given that the effective focal length is $f = 0.2\,\text{m}$, substitute the values:

$ \frac{1}{0.2} = \frac{2(1.5 - 1.2)}{1.2\,R}\,. $

Simplify the numerator:

$ \frac{1}{0.2} = \frac{2(0.3)}{1.2\,R} = \frac{0.6}{1.2\,R}\,. $

Since

$ \frac{0.6}{1.2} = 0.5, $

the equation becomes:

$ 5 = \frac{0.5}{R}\,. $

Solving for $R$:

$ R = \frac{0.5}{5} = 0.10\,\text{m}\,. $

Thus, the radius of curvature of the curved surface of the lens is $0.10\,\text{m}$.

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

For a thin plano-convex lens, one surface is flat, meaning $R_2 = \infty$ and $\frac{1}{R_2} = 0.$ The formula reduces to:

$ \frac{1}{f} = \left(\frac{n}{n'} - 1\right)\frac{1}{R_1} $

Thus, the focal length is:

$ f = \frac{R_1}{\left(\frac{n}{n'} - 1\right)} $

For the First Lens (in Air)

Refractive index of lens, $n = 1.5$

Surrounding medium (air), $n' = 1$

Radius of curvature, $R_1 = 2 \text{ cm}$

Substitute into the formula:

$ f_1 = \frac{2}{\left(\frac{1.5}{1} - 1\right)} = \frac{2}{(1.5 - 1)} = \frac{2}{0.5} = 4 \text{ cm} $

For the Second Lens (in Liquid)

Refractive index of lens, $n = 1.5$

Surrounding medium (liquid), $n' = 1.2$

Radius of curvature, $R_1 = 3 \text{ cm}$

Substitute into the formula:

$ f_2 = \frac{3}{\left(\frac{1.5}{1.2} - 1\right)} $

First, calculate the term inside the parentheses:

$ \frac{1.5}{1.2} = 1.25 \quad \Rightarrow \quad 1.25 - 1 = 0.25 $

Now, compute $f_2$:

$ f_2 = \frac{3}{0.25} = 12 \text{ cm} $

Ratio of the Focal Lengths

$ \frac{f_1}{f_2} = \frac{4}{12} = \frac{1}{3} $

Thus, the ratio of $f_1$ to $f_2$ is:

$ 1 : 3 $

When a concave mirror is dipped in a liquid, its focal length is still determined by the geometry of the mirror. For a spherical concave mirror, the focal length is given by

$ f = \frac{R}{2} $

where $ R $ is the radius of curvature of the mirror. This derivation comes solely from geometrical considerations and the law of reflection, which states that the angle of incidence equals the angle of reflection. Since reflection is independent of the medium in which the mirror is immersed, the focal length remains unchanged.

Thus, even when the mirror is in a liquid of refractive index $ \mu $, its focal length is still

$ f. $

The correct answer is Option B: $ f $.

$ \textbf{Step 1: Symmetry at Minimum Deviation} $

In a prism, when the deviation is minimum, the path of light is symmetric. This means that the angle of incidence ($i$) is equal to the angle of emergence ($i'$), and the light inside the prism makes equal angles with the prism faces. If the prism angle is $A$, then the refracted angle at each interface is:

$ r = \frac{A}{2} $

$ \textbf{Step 2: Relating Deviation to the Angles} $

The formula for the deviation ($D$) in a prism is given by:

$ D = i + i' - A $

At minimum deviation, $i = i'$, so:

$ D_{\text{min}} = 2i - A $

The special condition given is that the minimum deviation is equal to the prism angle:

$ D_{\text{min}} = A $

Thus:

$ A = 2i - A \quad \Longrightarrow \quad 2i = 2A \quad \Longrightarrow \quad i = A $

$ \textbf{Step 3: Applying Snell's Law} $

At the first surface, Snell's law gives:

$ \sin i = \mu \sin r $

Substitute the values $i = A$ and $r = \frac{A}{2}$:

$ \sin A = \mu \sin\frac{A}{2} $

Given that the refractive index $\mu = \sqrt{3}$, we have:

$ \sin A = \sqrt{3} \sin\frac{A}{2} $

$ \textbf{Step 4: Solving the Equation} $

Recall the double-angle formula for sine:

$ \sin A = 2 \sin\frac{A}{2} \cos\frac{A}{2} $

Substitute this into the previous equation:

$ 2 \sin\frac{A}{2} \cos\frac{A}{2} = \sqrt{3} \sin\frac{A}{2} $

Assuming $\sin\frac{A}{2} \neq 0$, we can divide both sides by $\sin\frac{A}{2}$:

$ 2 \cos\frac{A}{2} = \sqrt{3} $

Solve for $\cos\frac{A}{2}$:

$ \cos\frac{A}{2} = \frac{\sqrt{3}}{2} $

Since:

$ \cos 30^\circ = \frac{\sqrt{3}}{2} $

It follows:

$ \frac{A}{2} = 30^\circ \quad \Longrightarrow \quad A = 60^\circ $

$ \textbf{Final Answer:} $

The angle of the prism is $60^\circ$.

We can derive the lateral shift (d) by considering the geometry of a ray entering a parallel-sided slab. Here’s a step‐by‐step explanation:

The slab has a thickness $h$ (measured perpendicular to the surfaces). When the ray enters the glass from air at angle $i$ (with respect to the normal), it refracts inside and makes an angle $r$.

Inside the slab, the ray travels a distance

$L = \frac{h}{\cos r}$

because the vertical component of the path must cover a distance $h$.

The ray inside the slab is shifted laterally (parallel to the surface). Its lateral displacement over the distance $L$ is

$L \sin r = \frac{h \sin r}{\cos r}.$

However, the emergent ray is parallel to the incident ray but not collinear. By constructing the appropriate geometry (extending the emergent ray backwards until it meets the incident ray's extension at the top interface), you find that the lateral shift between the incident ray and the emergent ray is given by

$d = \frac{h \sin(i - r)}{\cos r}.$

This expression correctly accounts for both the refraction inside the slab and the geometry of the displacement.

Thus, the correct answer is:

Option D: $\frac{h \sin (i-r)}{\cos r}.$

We wish to find the distance from the point where the ray joining the object (O) and its image (I) meets the spherical surface (P) to the object. The given condition is that

  $PO = PI,$

meaning that P is the midpoint of OI.

Let’s work through the steps:

Setup the coordinate system and geometry:

 • Choose the vertex of the spherical surface (the “pole”) as the origin (x = 0).

 • Let the optical axis be the x‑axis.

 • Since the object is in air, place O at x = –u (with u > 0).

 • Since the real image is formed inside the glass, place I at x = v (with v > 0).

 • The spherical surface has radius of curvature $R$ and its center of curvature is in the glass. Thus, we take the center of the sphere as $C=(R,0)$.

 • The vertex (the point on the sphere closest to the object) is then located at x = 0.

Position of P:

 • The ray joining O and I is along the x‑axis. Its midpoint is

  $x_P = \frac{-u + v}{2}.$

 • However, since the physical part of the spherical surface is the convex cap met first by the incident light, the ray in fact meets the surface at the vertex, which is at x = 0.

 • Equating the midpoint to 0, we have

  $\frac{-u + v}{2}=0 \quad \Longrightarrow \quad v = u.$

 • Thus, the object and its image are equidistant from the vertex, and

  $PO = |0 - (-u)| = u.$

Relate u and R using the refraction at a spherical surface:

 • The formula for refraction at a spherical surface (using the appropriate sign–convention) is

  $\frac{n_1}{u} + \frac{n_2}{v} = \frac{n_2 - n_1}{R},$

  where for our problem

   – $n_1 = 1$ (air)

   – $n_2 = 1.5$ (glass).

 • With $v = u$, substitute in:

  $\frac{1}{u} + \frac{1.5}{u} = \frac{0.5}{R} \quad \Longrightarrow \quad \frac{2.5}{u} = \frac{0.5}{R}.$

 • Solving for u:

  $u = \frac{2.5\,R}{0.5} = 5R.$

Conclusion:

 • Since $PO = u$, we have

  $PO = 5R.$

Thus, the distance $PO$ equals $5R$, which is Option A.

$ \begin{aligned} & \frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{2}{\mathrm{f}_{\mathrm{L}}}-\frac{1}{\mathrm{f}_{\mathrm{m}}} \\\\ & \mathrm{f}_{\mathrm{m}}=-\frac{\left|\mathrm{R}_2\right|}{2} \\\\ & \frac{1}{\mathrm{f}_{\mathrm{L}}}=\left(\frac{\mu_2}{\mu_1}-1\right)\left(\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}\right) \end{aligned} $

Using lens maker formula, ${1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

Note : here, we will take air as medium for all three lenses so ${\mu _1} = 1$

${P_1} = {1 \over {{f_1}}} = \left( {{4 \over 3} - 1} \right)\left( {{1 \over \infty } - {1 \over { - {R_1}}}} \right)$

${P_1} = {1 \over 3}\left( {{1 \over {{R_1}}}} \right)$ ..... (1)

${P_2} = {1 \over {{f_2}}} = \left( {{3 \over 2} - 1} \right)\left( {{1 \over { - {R_1}}} - {1 \over { - {R_2}}}} \right)$

$ \Rightarrow {P_2} = {{ - 1} \over 2}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$ .... (2)

${P_3} = {1 \over {{f_3}}} = \left( {{4 \over 3} - 1} \right)\left( {{1 \over { - {R_2}}} - {1 \over \infty }} \right)$

$ \Rightarrow {P_3} = {{ - 1} \over 3}{1 \over {{R_2}}}$ ...... (3)

So, ${P_{eq}} = {1 \over {3{R_1}}} - {1 \over 2}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) - {1 \over {3{R_2}}}$

$ = {1 \over 3}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) - {1 \over 2}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

$ = \left( {{1 \over 3} - {1 \over 2}} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

${P_{eq}} = {{ - 1} \over 6}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

We know, for a combination of a lens and a mirror,

${P_{comb}} = 2{P_l} + {P_m}$ ...... (1)

Also using lens maker formula,

${1 \over {{f_l}}} = (\mu - 1)\left( {{1 \over R} - {1 \over { - R}}} \right) = (\mu - 1){2 \over R}$

So using eq. (i),

$ - {1 \over {{f_{comb}}}} = 2\left( {{1 \over {{f_l}}}} \right) - {1 \over {{f_m}}}$

$ \Rightarrow - {1 \over {{f_{comb}}}} = {{4(\mu - 1)} \over R} - {1 \over {\left( { - {R \over 2}} \right)}}$

$ \Rightarrow - {1 \over {{f_{comb}}}} = {{4(\mu - 1)} \over R} + {2 \over R}$

$ \Rightarrow - {1 \over {{f_{comb}}}} = {2 \over R}(2\mu - 2 + 1) = {2 \over R}(2\mu - 1)$

$ \Rightarrow {f_{comb}} = {{ - R} \over {2(2\mu - 1)}}$

$ \Rightarrow {R_{comb}} = 2{f_{comb}} = {{ - 2R} \over { - 2(2\mu - 1)}} = {{ - R} \over {2\mu - 1}}$

Hence, distance $ = {R \over {2\mu - 1}}$

$ \begin{aligned} & \frac{S-t}{2}+\frac{t}{2 n_0}=2 f \\ & S+t\left(\frac{1}{n_0}-1\right)=4 f \\ & S=4 f+\left(1-\frac{1}{n_0}\right) t \end{aligned} $

Also

$ \begin{aligned} & \frac{S-t}{2}+\frac{t}{2 n_0}=f \\ & S=2 f+\left(1-\frac{1}{n_0}\right) t \end{aligned} $

Parallel rays are focussed by convex lens and viceversa.

$\begin{aligned} \therefore \quad L_1 L_2 & =f_1+f_2 \\ & =10+15=25 \mathrm{~cm} \end{aligned}$

Let's analyze each statement carefully:

Statement (I): When an object is placed at the centre of curvature of a concave lens, image is formed at the centre of curvature of the lens on the other side.

To understand this statement, let's recall the nature of images formed by concave lenses. A concave lens (diverging lens) always forms images that are virtual, erect, and smaller than the object, regardless of the object's position. The term "centre of curvature" is typically associated with mirrors rather than lenses. Therefore, this statement is incorrect because a concave lens does not form a real image that could be said to be at the centre of curvature on the other side.

Statement (II): Concave lens always forms a virtual and erect image.

Now, considering the second statement, a concave lens (diverging lens) indeed always forms virtual, erect, and diminished images. This is a fundamental property of concave lenses.

Given this analysis:

The correct answer is Option A: Statement I is false but Statement II is true.

$\begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ & \frac{1}{v}-\frac{1}{-30}=\frac{1}{10} \\ & \frac{1}{v}=\frac{1}{10}-\frac{1}{30}=\frac{3-1}{30}=\frac{1}{15} \\ & v=15 \end{aligned}$

$\Rightarrow 30 \mathrm{~cm} \text { right of third }$

The critical angle is the angle of incidence at which light is refracted along the boundary, meaning the angle of refraction is $90^{\circ}$. The relationship between the critical angle and the refractive indices of two media can be understood using Snell's Law, given as:

$ n_1 \sin(\theta_c) = n_2 \sin(90^{\circ}) $

Here, $n_1$ is the refractive index of the first medium, $n_2$ is the refractive index of the second medium, and $\theta_c$ is the critical angle. Since $\sin(90^{\circ}) = 1$, the equation simplifies to:

$ n_1 \sin(\theta_c) = n_2 $

Given that the critical angle $\theta_c$ is $45^{\circ}$, we have:

$ \sin(45^{\circ}) = \frac{\sqrt{2}}{2} $

Substituting this value in the simplified Snell's Law equation, we get:

$ n_1 \left( \frac{\sqrt{2}}{2} \right) = n_2 $

Therefore,

$ n_1 = n_2 \cdot \frac{2}{\sqrt{2}} $

$ n_1 = n_2 \cdot \sqrt{2} $

Hence, the ratio of the refractive indices of the first and second media is:

$ n_1 : n_2 = \sqrt{2} : 1 $

The correct answer is therefore:

Option D

$\sqrt{2}: 1$

To find the refractive index of the material of a thin convex lens, we can make use of the Lensmaker's Formula. The Lensmaker's formula is given by:

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

where

• $f$ is the focal length of the lens,
• $\mu$ is the refractive index of the material of the lens,
• $R_1$ is the radius of curvature of the first surface (convex side, positive),
• $R_2$ is the radius of curvature of the second surface (concave side, negative for convex lens).

Given in the question, the radii of curvature are $15 \, \mathrm{cm}$ and $30 \, \mathrm{cm}$ respectively, and the focal length $f = 20\, \mathrm{cm}$.

It's important to pay attention to the signs of the radii of curvature according to the lens maker's convention. For convex lenses, $R_1$ is positive and $R_2$ is negative; however, since the problem doesn't specify which curvature corresponds to which side in relation to the direction of light travel, we'll assume the light travels from left to right: thus, $R_1 = +15 \, \mathrm{cm}$ and $R_2 = -30 \, \mathrm{cm}$.

Substituting the given values into the Lensmaker's Formula, we get:

$\frac{1}{20} = (\mu - 1) \left( \frac{1}{15} - \frac{1}{-30} \right) $

$\frac{1}{20} = (\mu - 1) \left( \frac{1}{15} + \frac{1}{30} \right) $

$\frac{1}{20} = (\mu - 1) \left( \frac{2 + 1}{30} \right) $

$\frac{1}{20} = (\mu - 1) \left( \frac{3}{30} \right) $

$\frac{1}{20} = (\mu - 1) \left( \frac{1}{10} \right) $

$\frac{1}{20} = \frac{\mu - 1}{10}$

Now, solve for $\mu$:

$\mu - 1 = \frac{10}{20} $

$\mu - 1 = 0.5 $

$\mu = 1.5$

Hence, the refractive index of the material of the lens is 1.5, which corresponds to Option B.

When we have a combination of identical lenses in contact, the effective power ($P_{\text{eff}}$) of the combination can be calculated as the sum of the powers of all the individual lenses. This is because the lenses are in direct contact, and their powers effectively add up.

Given that the effective power of a combination of 5 identical convex lenses is $25 \mathrm{D}$, we can use the formula for the effective power of the combination:

$P_{\text{eff}} = nP$

where:

• $P_{\text{eff}}$ is the effective power of the combination,
• $n$ is the number of lenses, and
• $P$ is the power of each individual lens.

Given $n = 5$ and $P_{\text{eff}} = 25 \mathrm{D}$, we can solve for $P$, the power of each lens:

$25 \mathrm{D} = 5P$

Dividing both sides by 5:

$P = \frac{25 \mathrm{D}}{5} = 5 \mathrm{D}$

The power of a lens ($P$) is related to its focal length ($f$) by the equation:

$P = \frac{1}{f}$

where $P$ is in diopters (D) and $f$ is in meters. Thus:

$5 \mathrm{D} = \frac{1}{f}$

Solving for $f$ gives:

$f = \frac{1}{5} \text{ meters} = \frac{1}{5} \times 100 \text{ cm} = 20 \text{ cm}$

Therefore, the focal length of each of the convex lenses is 20 cm. So, the correct answer is Option B: 20 cm.

First, let's understand the relationship between the object distance ($u$), the image distance ($v$), and the focal length ($f$) of a convex lens, which is given by the lens formula:

Now, to find the error in the focal length ($\Delta f$) due to the errors in the measurements of $u$ and $v$ ($\Delta u$ and $\Delta v$, respectively), we have to differentiate the lens formula with respect to $u$ and $v$, keeping in mind the propagation of error.

By differentiating both sides of the lens formula with respect to $v$ and $u$, and also considering the negative reciprocal relation (given $f$ is a constant for a specific lens), we have:

Rearranging this equation to find $\Delta f$, we get:

Therefore, the correct option showing the error in the measurement of the focal length of the convex lens, taking into account the least counts ($\Delta u$ and $\Delta v$) of the measuring scales for the position of the object ($u$) and for the position of the image ($v$), is:

Option C: $f^2\left[\frac{\Delta \mathrm{u}}{\mathrm{u}^2}+\frac{\Delta \mathrm{v}}{\mathrm{v}^2}\right]$

$\begin{aligned} & \mu=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\ & \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\ & \sin \left(\frac{\pi}{2}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_m}{2}\right) \\ & \frac{\pi}{2}-\frac{A}{2}=\frac{A}{2}+\frac{\delta m}{2} \\ & \delta_m=\pi-2 A \end{aligned}$

$\begin{aligned} & \mathrm{m}=2=\frac{-v}{u} \\ & 2=\frac{-(15-u)}{-u} \\ & 2 u=15-u \\ & 3 u=15 \Rightarrow u=5 \mathrm{~cm} \\ & v=15-u=15-5=10 \mathrm{~cm} \\ & \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ & =\frac{1}{10}+\frac{1}{(-5)}=\frac{1-2}{10}=\frac{-1}{10} \\ & f=-10 \mathrm{~cm} \end{aligned}$

Given $\mathrm{R}=30 \mathrm{~cm}$

$\mathrm{f}=\mathrm{R} / 2=+15 \mathrm{~cm}$

Magnification $(\mathrm{m})= \pm \frac{1}{2}$

For convex mirror, virtual image is formed for real object.

Therefore, $\mathrm{m}$ is +ve

$\begin{aligned} & \frac{1}{2}=\frac{f}{f-u} \\ & u=-15 \mathrm{~cm} \end{aligned}$

$\begin{aligned} & \mu_1=1.5 \\ & \mu_m=1.6 \\ & f_a=20 \mathrm{~cm} \\ & \text { As } \frac{f_m}{f_a}=\frac{\left(\mu_1-1\right) \mu_m}{\left(\mu_1-\mu_m\right)} \\ & \frac{f_m}{20}=\frac{(1.5-1) 1.6}{(1.5-1.6)} \\ & f_m=-160 \mathrm{~cm} \end{aligned}$

$\begin{aligned} & \cot \frac{\mathrm{A}}{2}=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right)}{\sin \frac{\mathrm{A}}{2}} \\ & \Rightarrow \cos \frac{\mathrm{A}}{2}=\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right) \\ & \frac{\mathrm{A}+\delta_{\min }}{2}=\frac{\pi}{2}-\frac{\mathrm{A}}{2} \\ & \delta_{\min }=\pi-2 \mathrm{~A} \end{aligned}%$

Spherometer can be used to measure curvature of surface.

As amount of energy incident on cell is same so current will remain same.