
A luminous point object is moving along the principal axis of a concave mirror of focal length 12 cm towards it. When its distance from the mirror is 20 cm its velocity is 4 cm/s. The velocity of the image in cm/s at that instant is:
A. 6, towards the mirror
B. 6, away from the mirror
C. 9, away from the mirror
D. 9, towards the mirror
Answer
216.3k+ views
Hint: A concave mirror is a type of spherical mirror which has an inner reflecting surface and its outer part is painted. It is capable of converging all the light rays that are incident on it, at a single point.
Formula Used:
Mirror formula is a formula that gives the relation between object distance, image distance and the focal length of the mirror. Mathematically, mirror formula is written as
\[\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\]
Complete step by step solution:
Given that the focal length of a concave mirror is, f=-12 (Since focal length for a concave mirror is always negative).
Also given that the object distance will be, u=-20 cm
Given velocity of the object is 4 cm/s.
Using mirror formula and substituting all the values in the equation and solving, we get
\[\dfrac{1}{v} = - \dfrac{1}{{12}} - \dfrac{1}{{( - 20)}}\]
\[\Rightarrow \dfrac{1}{v} = - \dfrac{1}{{12}} + \dfrac{1}{{20}}\]
\[\Rightarrow \dfrac{1}{v} = \dfrac{{3 - 5}}{{60}}\]
\[\Rightarrow \dfrac{1}{v} = - \dfrac{1}{{30}}\]
$\Rightarrow v=-30\,cm$
Velocity of the image is given by the formula,
\[\dfrac{{dv}}{{dt}} = - (\dfrac{{{v^2}}}{{{u^2}}})\dfrac{{du}}{{dt}}\]
\[\Rightarrow \dfrac{{dv}}{{dt}} = - (\dfrac{{{{( - 30)}^2}}}{{{{(20)}^2}}})4\]
\[\Rightarrow \dfrac{{dv}}{{dt}} = - (\dfrac{{900}}{{400}})4\]
\[\therefore \dfrac{{dv}}{{dt}} = - 9\,cm{s^{ - 1}}\]
Therefore, the velocity of the image will be 9, away from the mirror.
Hence, Option C is the correct answer
Note: It is important to remember that concave mirror is also known as converging mirror. This is because, when light rays fall on the surface of the mirror, all the right rays get reflected and converge at a single point, making an image at that point. The nature of the image formed by a concave mirror depends on the position of the object. If it is placed closer to the mirror, the mirror forms a virtual and magnified image. If it is placed at a distance from the mirror, it forms a real and diminished image.
Formula Used:
Mirror formula is a formula that gives the relation between object distance, image distance and the focal length of the mirror. Mathematically, mirror formula is written as
\[\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\]
Complete step by step solution:
Given that the focal length of a concave mirror is, f=-12 (Since focal length for a concave mirror is always negative).
Also given that the object distance will be, u=-20 cm
Given velocity of the object is 4 cm/s.
Using mirror formula and substituting all the values in the equation and solving, we get
\[\dfrac{1}{v} = - \dfrac{1}{{12}} - \dfrac{1}{{( - 20)}}\]
\[\Rightarrow \dfrac{1}{v} = - \dfrac{1}{{12}} + \dfrac{1}{{20}}\]
\[\Rightarrow \dfrac{1}{v} = \dfrac{{3 - 5}}{{60}}\]
\[\Rightarrow \dfrac{1}{v} = - \dfrac{1}{{30}}\]
$\Rightarrow v=-30\,cm$
Velocity of the image is given by the formula,
\[\dfrac{{dv}}{{dt}} = - (\dfrac{{{v^2}}}{{{u^2}}})\dfrac{{du}}{{dt}}\]
\[\Rightarrow \dfrac{{dv}}{{dt}} = - (\dfrac{{{{( - 30)}^2}}}{{{{(20)}^2}}})4\]
\[\Rightarrow \dfrac{{dv}}{{dt}} = - (\dfrac{{900}}{{400}})4\]
\[\therefore \dfrac{{dv}}{{dt}} = - 9\,cm{s^{ - 1}}\]
Therefore, the velocity of the image will be 9, away from the mirror.
Hence, Option C is the correct answer
Note: It is important to remember that concave mirror is also known as converging mirror. This is because, when light rays fall on the surface of the mirror, all the right rays get reflected and converge at a single point, making an image at that point. The nature of the image formed by a concave mirror depends on the position of the object. If it is placed closer to the mirror, the mirror forms a virtual and magnified image. If it is placed at a distance from the mirror, it forms a real and diminished image.
Recently Updated Pages
Wheatstone Bridge Explained: Working, Formula & Uses

Young’s Double Slit Experiment Derivation Explained

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

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

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Average and RMS Value in Electrical Circuits

