The image formed by convex mirror of focal length $30 \mathrm{cm}$ is the quarter of size of
the object. Then the distance of the object from the mirror is,
A. $30 \mathrm{cm}$
B. $90 \mathrm{cm}$
C. $120 \mathrm{cm}$
D. $60 \mathrm{cm}$
Answer
Verified
116.1k+ views
Hint Optics is the part of material science that reviews the conduct and properties of light, incorporating its communications with issues and the development of instruments that utilize or distinguish it. Optics as a rule portrays the conduct of obvious, bright, and infrared light. We must keep this in mind before attempting any such question. To proceed we must put the formula for calculation of focal length and formula for magnification.
Complete step by step answer:
From the given question, we know that, the focal length of the convex mirror is $\mathrm{f}=30 \mathrm{cm}$ and that the magnification of the image is $\text{m}=\dfrac{1}{4}$. We have to calculate the object distance that is denoted by u. According to the sign convention, we know that the measurements along the direction of light are taken as positive and that opposite to the light are taken as negative. Therefore, we derive that the transverse measurement above the principal axis is taken as positive and that below the principal axis is taken as negative. Therefore, $f=+30 \mathrm{cm}$ and $\mathrm{m}=+1 / 4=+0.25$.
Using the formula, $\mathrm{m}=\dfrac{\mathrm{f}}{\mathrm{f}-\mathrm{u}}$
$\Rightarrow \text{mf}-\text{mu}=\text{f}$
$\Rightarrow \text{u}=\dfrac{\text{mf}-\text{f}}{\text{m}}=\dfrac{\text{m}-1}{\text{m}}\text{f}=\dfrac{0.25-1}{0.25}(30)=-90\text{cm}$
The minus sign prevails due to the assumption in direction but because we are looking into magnitudes only.
The correct answer is Option B.
Note: We must keep in mind that in optics, the sign related with the result portrays the direction only but to answer such questions, we only consider magnitudes.
Complete step by step answer:
From the given question, we know that, the focal length of the convex mirror is $\mathrm{f}=30 \mathrm{cm}$ and that the magnification of the image is $\text{m}=\dfrac{1}{4}$. We have to calculate the object distance that is denoted by u. According to the sign convention, we know that the measurements along the direction of light are taken as positive and that opposite to the light are taken as negative. Therefore, we derive that the transverse measurement above the principal axis is taken as positive and that below the principal axis is taken as negative. Therefore, $f=+30 \mathrm{cm}$ and $\mathrm{m}=+1 / 4=+0.25$.
Using the formula, $\mathrm{m}=\dfrac{\mathrm{f}}{\mathrm{f}-\mathrm{u}}$
$\Rightarrow \text{mf}-\text{mu}=\text{f}$
$\Rightarrow \text{u}=\dfrac{\text{mf}-\text{f}}{\text{m}}=\dfrac{\text{m}-1}{\text{m}}\text{f}=\dfrac{0.25-1}{0.25}(30)=-90\text{cm}$
The minus sign prevails due to the assumption in direction but because we are looking into magnitudes only.
The correct answer is Option B.
Note: We must keep in mind that in optics, the sign related with the result portrays the direction only but to answer such questions, we only consider magnitudes.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs
How to find Oxidation Number - Important Concepts for JEE
How Electromagnetic Waves are Formed - Important Concepts for JEE
Electrical Resistance - Important Concepts and Tips for JEE
Average Atomic Mass - Important Concepts and Tips for JEE
Chemical Equation - Important Concepts and Tips for JEE
Trending doubts
JEE Main 2025: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main Chemistry Question Paper with Answer Keys and Solutions
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
Charging and Discharging of Capacitor
Other Pages
NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
Units and Measurements Class 11 Notes - CBSE Physics Chapter 1
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids