
Let the transverse magnification produced by a spherical mirror is \[m\]. Then for the same position of objective mirror the longitudinal magnification will be:
A. \[m\]
B. \[\sqrt m \]
C. \[{m^2}\]
D. \[\dfrac{1}{m}\]
Answer
218.7k+ views
Hint As we know lateral or transverse magnification are same and is given as \[m\]and we know the formula to calculate the longitudinal magnification and can be calculated through \[{m_{longitud}} = {\left( {{m_{trans}}} \right)^2}\].
Complete step by step answer As we know we are given as transverse magnification as, \[{m_{trans}} = m\]
And to calculate the longitudinal magnification as, \[{m_{longitud}} = {\left( {{m_{trans}}} \right)^2}\]
By substituting the transverse in formula we get, \[{m_{longitud}} = {m^2}\]
Therefore we get as \[{m^2}\]the longitudinal magnification.
Additional information Lateral magnification is referred to relative heights of objects to height of image in our ray tracings. Through transverse magnification we can calculate the ratio of heights of object and image and we can also calculate the velocities of both the image and object. And transverse magnification changes along optical axes whereas longitudinal magnification is not linear and 3D image is distorted through longitudinal magnification. The word "lateral" is appended above because it only applies to the dimensions of the objects perpendicular to the optical axis. As stated above, this is almost exclusively what we will be working with, let's take a moment to look at magnification in the cases of the plane reflector when the object arrow is parallel to the optical axis.
Note As we know about lateral magnification its formula is \[m = - \dfrac{v}{u}\], here we have negative sign but we always choose its magnitude but if we have negative sign in option then we have to take care of this negative sign and take signed value of velocities that is with direction. And the same is in the case of longitudinal magnification.
Complete step by step answer As we know we are given as transverse magnification as, \[{m_{trans}} = m\]
And to calculate the longitudinal magnification as, \[{m_{longitud}} = {\left( {{m_{trans}}} \right)^2}\]
By substituting the transverse in formula we get, \[{m_{longitud}} = {m^2}\]
Therefore we get as \[{m^2}\]the longitudinal magnification.
Additional information Lateral magnification is referred to relative heights of objects to height of image in our ray tracings. Through transverse magnification we can calculate the ratio of heights of object and image and we can also calculate the velocities of both the image and object. And transverse magnification changes along optical axes whereas longitudinal magnification is not linear and 3D image is distorted through longitudinal magnification. The word "lateral" is appended above because it only applies to the dimensions of the objects perpendicular to the optical axis. As stated above, this is almost exclusively what we will be working with, let's take a moment to look at magnification in the cases of the plane reflector when the object arrow is parallel to the optical axis.
Note As we know about lateral magnification its formula is \[m = - \dfrac{v}{u}\], here we have negative sign but we always choose its magnitude but if we have negative sign in option then we have to take care of this negative sign and take signed value of velocities that is with direction. And the same is in the case of longitudinal magnification.
Recently Updated Pages
Two discs which are rotating about their respective class 11 physics JEE_Main

A ladder rests against a frictionless vertical wall class 11 physics JEE_Main

Two simple pendulums of lengths 1 m and 16 m respectively class 11 physics JEE_Main

The slopes of isothermal and adiabatic curves are related class 11 physics JEE_Main

A trolly falling freely on an inclined plane as shown class 11 physics JEE_Main

The masses M1 and M2M2 M1 are released from rest Using class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

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

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

How to Convert a Galvanometer into an Ammeter or Voltmeter

