JEE Advanced · Physics · 24. Ray Optics

Consider a concave mirror and a convex lens (refractive index = $1.5$ ) of focal length $10$ $cm$ each, separated by a distance of $50$ $cm$ in air (refractive index = $1$ ), as shown in the figure. An object is placed at a distance of $15$ $cm$ from the mirror. Its erect image, formed by this combination, has magnification $M_{1}$ . When the set-up is kept in a medium of refractive index $\frac{7}{6}$ , the magnification becomes $M_{2}$ . The magnitude $|\frac{M_{2}}{M_{1}}|$ is

A 5
B 7
C 9
D 11

Verified Solution

Answer & Solution

Correct Answer

(B) 7

Step-by-step Solution

Detailed explanation

For reflection from a concave mirror,

\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} - \frac{1}{15} = \frac{- 1}{10}

\frac{1}{v} = \frac{1}{15} - \frac{1}{10} = \frac{- 1}{30}

∴ v = - 30

Magnification

(m_{1}) = - \frac{v}{u} = - 2

Now for refraction from lens,

\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{10} - \frac{1}{20} = \frac{1}{20}

∴

Magnification

(m_{2}) = \frac{v}{u} = - 1

∴ M_{1} = m_{1} m_{2} = 2

Now when the set-up is immersed in liquid, no effect for the image formed by mirror.
We have

(μ_{L} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = \frac{1}{10}

\Rightarrow (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = \frac{1}{5}

When lens is immersed in liquid,

\frac{1}{f_{l e n s}} = (\frac{μ_{L}}{μ_{S}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = \frac{2}{7} \times \frac{1}{5} = \frac{2}{35}

∴ \frac{1}{v} - \frac{1}{u} = \frac{1}{f_{L i q u i d}}

\Rightarrow \frac{1}{v} = \frac{2}{35} - \frac{1}{20} = \frac{8 - 7}{140} = \frac{1}{140}

∴

Magnification

= - \frac{140}{20} = - 7

∴ M_{2} = 2 \times 7 = 14

∴ |\frac{M_{2}}{M_{1}}| = 7

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

The electric field E is measured at a point

P (0,0, d)

, generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. List-I contains different relations between E and d. List-II describes different electric charge distributions, along with their locations. Match the functions in List-I with the related charge distributions in List-II.

	LIST -I		LIST -II
A)	is independent of d	P)	A point charge at the origin
B)	$E \propto \frac{1}{d}$	Q)	A small dipole with point charges $Q$ at $(0,0, l)$ and $-\operatorname{Qat}(0,0, l)$. Take $2 l \ll d$
C)	$E \propto \frac{1}{d^2}$	R)	An infinite line charge coincident with the $x$-axis, with uniform linear charge density $\lambda$
D)	$E \propto \frac{1}{d^3}$	S)	Two infinite wires carrying uniform linear charge density parallel to the $x$ - axis. The one along $(y-0, z=l)$ has a charge density $+\lambda$ and the one along $(y=0, z=-l)$ has a charge density $-\lambda$. Take $2 l \ll d$
		T)	Infinite plane charge coincident with the $x y$ - plane with uniform surface charge density

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