
Convex surface of the thin concavo-convex lens of refractive index 1.5 is silvered as shown. A small object kept in air at 30 cm left of the lens on its principal axis. The distance of the final image from mirror is

A. 20 cm
B. 30 cm
C. 10 cm
D. 15 cm
Answer
163.2k+ views
Hint:Here the refractive index of the thin concavo-convex lens is given and it is given that the convex surface of the lens is silvered. An object is kept at 30 cm left of the lens on its principal axis, we have to find the distance of the final image from the mirror. By applying lens makers formula and mirror formula, we can find the distance of the final image.
Formula used:
Lens makers formula:
$\dfrac{1}{f}=(\mu -1)\left( \dfrac{1}{{{r}_{1}}}-\dfrac{1}{{{r}_{2}}} \right)$
Where f= focal length, μ=refractive index, r1, r2=radius of curvature of both sides.
Mirror formula:
$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$
Where f=focal length, v=image distance and u=object distance.
Complete step by step solution:
Thin concavo-convex lens has refractive index 1.5. Its convex side is coated with silver. That is, it is now acting as a mirror not lens. Given a small object is kept in air at 30 cm left of the lens on its principal axis as shown in the figure.

We have to find the distance of the final image from the mirror.
On applying Lens makers formula, we get focal length of lens as:
We have equation as:
$\dfrac{1}{f}=(\mu -1)\left( \dfrac{1}{{{r}_{1}}}-\dfrac{1}{{{r}_{2}}} \right)$
On substituting values, we get:
$\dfrac{1}{f}=(1.5-1)\left( -\dfrac{1}{60}+\dfrac{1}{20} \right)$
On solving,
Focal length of lens, $f=60cm$
But we know that the convex side of a thin concavo-convex lens is silvered so it acts as a mirror. Now focal length of combination of lens and mirror is
$\dfrac{1}{{{f}_{eq}}}=\dfrac{1}{{{f}_{mirror}}}-\dfrac{2}{f}=-\dfrac{1}{10}-\dfrac{2}{60}=\dfrac{-2}{15}cm$
$\therefore {{f}_{eq}}=\dfrac{-15}{2}$
Now applying a mirror formula to find the distance of the final image.
That is,
$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=-\dfrac{2}{15}+\dfrac{1}{30}=-\dfrac{1}{10}cm$
That is, the distance of the final image from the mirror is 10 cm.
Therefore, the answer is option C.
Notes: In this kind of problem do not forget to apply sign convention. Here when we solve a problem, we use sign convention and we ignore the sign on the final answer because the sign says the position of the image with respect to the lens or mirror we are using and magnitude only matters.
Formula used:
Lens makers formula:
$\dfrac{1}{f}=(\mu -1)\left( \dfrac{1}{{{r}_{1}}}-\dfrac{1}{{{r}_{2}}} \right)$
Where f= focal length, μ=refractive index, r1, r2=radius of curvature of both sides.
Mirror formula:
$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$
Where f=focal length, v=image distance and u=object distance.
Complete step by step solution:
Thin concavo-convex lens has refractive index 1.5. Its convex side is coated with silver. That is, it is now acting as a mirror not lens. Given a small object is kept in air at 30 cm left of the lens on its principal axis as shown in the figure.

We have to find the distance of the final image from the mirror.
On applying Lens makers formula, we get focal length of lens as:
We have equation as:
$\dfrac{1}{f}=(\mu -1)\left( \dfrac{1}{{{r}_{1}}}-\dfrac{1}{{{r}_{2}}} \right)$
On substituting values, we get:
$\dfrac{1}{f}=(1.5-1)\left( -\dfrac{1}{60}+\dfrac{1}{20} \right)$
On solving,
Focal length of lens, $f=60cm$
But we know that the convex side of a thin concavo-convex lens is silvered so it acts as a mirror. Now focal length of combination of lens and mirror is
$\dfrac{1}{{{f}_{eq}}}=\dfrac{1}{{{f}_{mirror}}}-\dfrac{2}{f}=-\dfrac{1}{10}-\dfrac{2}{60}=\dfrac{-2}{15}cm$
$\therefore {{f}_{eq}}=\dfrac{-15}{2}$
Now applying a mirror formula to find the distance of the final image.
That is,
$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=-\dfrac{2}{15}+\dfrac{1}{30}=-\dfrac{1}{10}cm$
That is, the distance of the final image from the mirror is 10 cm.
Therefore, the answer is option C.
Notes: In this kind of problem do not forget to apply sign convention. Here when we solve a problem, we use sign convention and we ignore the sign on the final answer because the sign says the position of the image with respect to the lens or mirror we are using and magnitude only matters.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Charging and Discharging of Capacitor

Instantaneous Velocity - Formula based Examples for JEE

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

In which of the following forms the energy is stored class 12 physics JEE_Main

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Other Pages
Three mediums of refractive indices mu 1mu 0 and mu class 12 physics JEE_Main

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks

NEET Marks vs Rank 2024|How to Calculate?
