JEE Advanced · Physics · 24. Ray Optics

A thin convex lens is made of two materials with refractive indices $n_{1}$ and $n_{2},$ as shown in figure. The radius of curvature of the left and right spherical surfaces are equal. $f$ is the focal length of the lens when $n_{1} = n_{2} = n .$ The focal length is $f + Δ f$ when $n_{1} = n$ and $n_{2} = n + Δ n .$ Assuming $Δ n < < (n - 1)$ and $1 < n < 2,$ the correct statement(s) is/are:

A For $n = 1.5, Δ n = 10^{- 3}$ and $f = 20 c m,$ the value of $|Δ f|$ will be $0.02 c m$ (round off to $2^{n d}$ decimal place)
B The relation between $\frac{Δ f}{f}$ and $\frac{Δ n}{n}$ remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature.
C $|\frac{Δ f}{f}| < |\frac{Δ n}{n}|$
D If $\frac{Δ n}{n} < 0$ then $\frac{Δ f}{f} > 0$

Verified Solution

Answer & Solution

Correct Answer

(A) For $n = 1.5, Δ n = 10^{- 3}$ and $f = 20 c m,$ the value of $|Δ f|$ will be $0.02 c m$ (round off to $2^{n d}$ decimal place)

Step-by-step Solution

Detailed explanation

Given that the lens is made of two plano-convex lenses. Let it be

L_{1}

L_{2}

Let focal length of

L_{1} = f_{1},

An focal length of

L_{2} = f_{2}

∴

Using lens maker's formula,

\frac{1}{f_{1}} = (n_{1} - 1) (\frac{1}{R} - \frac{1}{\infty}),

and

\frac{1}{f_{2}} = (n_{2} - 1) (\frac{1}{\infty} - \frac{1}{(- R)})

\Rightarrow \frac{1}{f_{1}} = \frac{n_{1} - 1}{R}

\Rightarrow \frac{1}{f_{2}} = \frac{n_{2} - 1}{R}

Now, for the first situation, given that,

n_{1} = n_{2} = n

and focal length of combination is

‘ f ’

\Rightarrow \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{2 (n - 1)}{R}

....(1)
In the second situation,

n_{1} = n,

n_{2} = n + Δ n

and focal length=

f + Δ f

\Rightarrow \frac{1}{f + Δ f} = \frac{n - 1}{R} + \frac{n + Δ n - 1}{R}

\Rightarrow \frac{1}{f + Δ f} = \frac{2 n + Δ n - 2}{R}

.....(2)
Nos dividing equation (1)/(2) we get

\frac{f + Δ f}{f} = \frac{2 (n - 1)}{2 n + Δ n - 2} \Rightarrow 1 + \frac{Δ f}{f} = \frac{(2 n - 2 + Δ n) - Δ n}{2 n + Δ n - 2}

\Rightarrow 1 + \frac{Δ f}{f} = 1 - \frac{Δ n}{2 n + Δ n - 2} \Rightarrow \frac{Δ f}{f} = \frac{- Δ n}{2 n + Δ n - 2}

Now, as

' Δ n^{'}

is very small

\Rightarrow \frac{Δ f}{f} = \frac{- Δ n}{2 (n - 1)} \dots . . (3)

Now, option

(A)

n = 1.5, Δ n = 10^{- 3} a n d f = 20 c m

\Rightarrow | Δ f | = \frac{f \cdot (Δ n)}{2 (n - 1)} = \frac{20 \times 10^{- 3}}{2 \times (1.5 - 1)} = 0.02 c m \Rightarrow (A)

is correct.
Option

(B)

The relation between

\frac{Δ f}{f}

\frac{Δ n}{n}

is dependent of

‘ R ’

\Rightarrow

Remain unchanged for convex or concave surfaces

\Rightarrow (B)

is correct.
Option

(C)

n < 2 \Rightarrow 2 (n - 1) < n

\Rightarrow \frac{Δ f}{f} = \frac{1}{2} |\frac{Δ n}{n - 1}| > \frac{Δ n}{n}

\Rightarrow \frac{Δ f}{f} > \frac{Δ n}{n} \Rightarrow (C)

is incorrect.
Option

(D)

\frac{Δ n}{n} < 0,

then

\frac{Δ f}{f} > 0,

using equation

(3)

\Rightarrow (D)

is correct.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

List-I shows four configurations, each consisting of a pair of ideal electric dipoles. Each dipole has a dipole moment of magnitude $p$, oriented as marked by arrows in the figures. In all the configurations the dipoles are fixed such that they are at a distance $2 r$ apart along the $x$ direction. The midpoint of the line joining the two dipoles is $X$. The possible resultant electric fields $\vec{E}$ at $X$ are given in List-II. Choose the option that describes the correct match between the entries in List-I to those in List-II.

		List-II
$(P)$	$(1)$	$\vec{E}=0$
$(Q)$	$(2)$	$\vec{E}=-\frac{p}{2 \pi \epsilon_0 \mathrm{r}^3} \hat{\mathrm{j}}$
$(R)$	$(3)$	$\vec{E}=-\frac{p}{4 \pi \epsilon_0 \mathrm{r}^3}(\hat{\mathrm{i}}-\hat{\mathrm{j}})$

$(S)$	$(4)$	$\vec{E}=\frac{p}{4 \pi \epsilon_0 \mathrm{r}^3}(2 \hat{\mathrm{i}}-\hat{\mathrm{j}})$
	$(5)$	$\vec{E}=\frac{p}{\pi \epsilon_0 \mathrm{r}^3} \hat{\mathrm{i}}$

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		List-II
\((P)\)	\((1)\)	\(\vec{E}=0\)
\((Q)\)	\((2)\)	\(\vec{E}=-\frac{p}{2 \pi \epsilon_0 \mathrm{r}^3} \hat{\mathrm{j}}\)
\((R)\)	\((3)\)	\(\vec{E}=-\frac{p}{4 \pi \epsilon_0 \mathrm{r}^3}(\hat{\mathrm{i}}-\hat{\mathrm{j}})\)

\((S)\)	\((4)\)	\(\vec{E}=\frac{p}{4 \pi \epsilon_0 \mathrm{r}^3}(2 \hat{\mathrm{i}}-\hat{\mathrm{j}})\)
	\((5)\)	\(\vec{E}=\frac{p}{\pi \epsilon_0 \mathrm{r}^3} \hat{\mathrm{i}}\)