
The focal length of a concave mirror is $50{\text{ cm}}$ where an object is to be placed so that its image is two times magnified, real and inverted:
A. $75{\text{ cm}}$
B. $72{\text{ cm}}$
C. ${\text{63 cm}}$
D. ${\text{50 cm}}$
Answer
124.5k+ views
Hint A concave mirror is a spherical mirror whose reflecting surface is its bent surface. There is a relation between the focal length of a mirror $f$, distance of an object $u$ and distance of the image formed $v$, known as Mirror formula and given by $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ .
Linear magnification of a mirror is the ratio of height of the image formed by the mirror to the height of the object and which is given by $m = \dfrac{{ - v}}{u}$ where $v$ is the distance of the image and $u$ is the distance of the object. If an object is magnified by two times and is inverted then its linear magnification $m = - 2$ .
Complete step by step answer
Let us first know about a concave mirror.
A concave mirror is a spherical mirror whose reflecting surface is its bent surface.
As given in the question that the focal length of a concave mirror, $f = 50{\text{ cm}}$ .
We are asked to find the distance of the object $u$ such that the image formed is two times magnified, real and inverted.
If an object is magnified by two times and is inverted then its linear magnification $m = - 2$ .
Let us discuss the linear magnification of a mirror.
Linear magnification of a mirror is the ratio of height of the image formed by the mirror to the height of the object and which is given by $m = \dfrac{{ - v}}{u}$ where $v$ is the distance of the image and $u$ is the distance of the object.
So, according to the question,
$\dfrac{{ - v}}{u} = - 2$ which simplifies to $v = 2u$
As we know that there is a relation between the focal length of a mirror $f$, distance of an object $u$ and distance of the image formed $v$, known as Mirror formula and given by $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ .
So, substituting the values in the mirror formula we have
$\dfrac{1}{{50}} = \dfrac{1}{{2u}} + \dfrac{1}{u} = \dfrac{3}{{2u}}$
On simplifying we have
$u = 75cm$
Therefore, the object is to be placed at distance $75{\text{ cm}}$ from the mirror so that its image is two times magnified, real and inverted.
Hence, option A is correct
Note Here, we take negative sign for the magnification because the image formed is inverted i.e. below the central axis of the mirror.
Distances of any point are measured along the central axis by convention, as the direction towards the object in mirror is taken as positive and away from the object is taken as negative. Hence the radius of curvature and the focal length are negative for convex mirrors and positive for concave mirrors.
Linear magnification of a mirror is the ratio of height of the image formed by the mirror to the height of the object and which is given by $m = \dfrac{{ - v}}{u}$ where $v$ is the distance of the image and $u$ is the distance of the object. If an object is magnified by two times and is inverted then its linear magnification $m = - 2$ .
Complete step by step answer
Let us first know about a concave mirror.
A concave mirror is a spherical mirror whose reflecting surface is its bent surface.
As given in the question that the focal length of a concave mirror, $f = 50{\text{ cm}}$ .
We are asked to find the distance of the object $u$ such that the image formed is two times magnified, real and inverted.
If an object is magnified by two times and is inverted then its linear magnification $m = - 2$ .
Let us discuss the linear magnification of a mirror.
Linear magnification of a mirror is the ratio of height of the image formed by the mirror to the height of the object and which is given by $m = \dfrac{{ - v}}{u}$ where $v$ is the distance of the image and $u$ is the distance of the object.
So, according to the question,
$\dfrac{{ - v}}{u} = - 2$ which simplifies to $v = 2u$
As we know that there is a relation between the focal length of a mirror $f$, distance of an object $u$ and distance of the image formed $v$, known as Mirror formula and given by $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ .
So, substituting the values in the mirror formula we have
$\dfrac{1}{{50}} = \dfrac{1}{{2u}} + \dfrac{1}{u} = \dfrac{3}{{2u}}$
On simplifying we have
$u = 75cm$
Therefore, the object is to be placed at distance $75{\text{ cm}}$ from the mirror so that its image is two times magnified, real and inverted.
Hence, option A is correct
Note Here, we take negative sign for the magnification because the image formed is inverted i.e. below the central axis of the mirror.
Distances of any point are measured along the central axis by convention, as the direction towards the object in mirror is taken as positive and away from the object is taken as negative. Hence the radius of curvature and the focal length are negative for convex mirrors and positive for concave mirrors.
Recently Updated Pages
Young's Double Slit Experiment Step by Step Derivation

Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

The formula of the kinetic mass of a photon is Where class 12 physics JEE_Main

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Main Login 2045: Step-by-Step Instructions and Details

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Ideal and Non-Ideal Solutions Raoult's Law - JEE

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!
