
For magnification in spherical mirror object height is:
(A) Negative
(B) Positive
(C) For real images positive
(D) For virtual images negative
Answer
174k+ views
Hint If we want to magnify an image of an object, we use a spherical mirror called a concave mirror. Next, we need to recall the conventions which we use for making ray diagrams involving concave mirrors. The object height is always taken positive if the object is in the positive y-direction.
Complete step by step answer
Spherical mirrors: it is a part of a spherical surface mirror. It is of two types- one is concave and another one is convex. The bulged out part is known as convex mirror whereas it is thinner from the center and thicker at the edges then it is known as concave mirror.
Magnification is defined as the ratio of image height to an object height. It is denoted by ‘m’.
$m = \dfrac{{{h_i}}}{{{h_o}}}$ Where ${h_{i\,}}$ -image height; ${h_o}$ - object height; m- magnification.
It is used to increase the height of the object.
Real image: when an image is formed on the screen.
Virtual image: when an image appears to form on the screen.
Image gets shorter and longer depending upon mirrors, but the object is always placed above the principal axis.
So, it will always be positive. Therefore, option B is correct.
Note
If object height is taken as negative then it should be placed below the principal axis, but that is against the well-established convention, so it is unwise to think in that direction. Hence option A is considered wrong.
Complete step by step answer
Spherical mirrors: it is a part of a spherical surface mirror. It is of two types- one is concave and another one is convex. The bulged out part is known as convex mirror whereas it is thinner from the center and thicker at the edges then it is known as concave mirror.
Magnification is defined as the ratio of image height to an object height. It is denoted by ‘m’.
$m = \dfrac{{{h_i}}}{{{h_o}}}$ Where ${h_{i\,}}$ -image height; ${h_o}$ - object height; m- magnification.
It is used to increase the height of the object.
Real image: when an image is formed on the screen.
Virtual image: when an image appears to form on the screen.
Image gets shorter and longer depending upon mirrors, but the object is always placed above the principal axis.
So, it will always be positive. Therefore, option B is correct.
Note
If object height is taken as negative then it should be placed below the principal axis, but that is against the well-established convention, so it is unwise to think in that direction. Hence option A is considered wrong.
Recently Updated Pages
JEE Main 2025-26 Atoms and Nuclei Mock Test: Free Practice Online

JEE Main 2025-26: Dual Nature of Matter and Radiation Mock Test

JEE Main 2025-26 Electronic Devices Mock Test – Free Practice

JEE Main Mock Test 2025-26: Experimental Skills Chapter Online Practice

JEE Main 2025-26 Current Electricity Mock Test: Free Practice Online

JEE Main 2025-26 Rotational Motion Mock Test – Free Practice Online

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

Electric field due to uniformly charged sphere class 12 physics JEE_Main

JEE Main 2025: Derivation of Equation of Trajectory in Physics

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

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

Instantaneous Velocity - Formula based Examples for JEE

What is Hybridisation in Chemistry?

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Essential Derivations for CBSE Class 12 Physics: Stepwise & PDF Solutions

Electron Gain Enthalpy and Electron Affinity for JEE
