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QJEEM23Phd25d: JEE Main 2023 Physics - Class 12 JEEM23Phd25d Solution: Two objects A and B are placed at 15 cm and 25 cm from the pole …

Two objects A and B are placed at 15 cm and 25 cm from the pole in front of a concave mirror having radius of curvature 40 cm. The distance between images formed by the mirror is _______________.

Multiple Choice Options

  • A 60 cm
  • B 40 cm
  • C 160 cm
  • D 100 cm

Step-by-Step Verified Solution

JEE Main 2023 (Online) 1st February Evening Shift Physics - Geometrical Optics Question 76 English Explanation

Using Mirror formula

amp;1v+1u=1famp;1v=1f−1famp;v=ufu−f\begin{aligned}& \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\\\& \frac{1}{v}=\frac{1}{f}-\frac{1}{f} \\\\& v=\frac{u f}{u-f}\end{aligned}

For object A,ui=−15 cm,f=−20 cm, v1=\mathrm{A}, \mathrm{u}_{\mathrm{i}}=-15 \mathrm{~cm}, \mathrm{f}=-20 \mathrm{~cm}, \mathrm{~v}_1= ?

amp;v1=u1fu1−f=(−15)(−20)(−15)−(20)=+3005amp;v1=+60 cm(+ve,virtual)\begin{aligned}& \mathrm{v}_1=\frac{\mathrm{u}_1 \mathrm{f}}{\mathrm{u}_1-\mathrm{f}}=\frac{(-15)(-20)}{(-15)-(20)}=\frac{+300}{5} \\\\& \mathrm{v}_1=+60 \mathrm{~cm}(+ ve, virtual)\end{aligned}

For object B,u2=−25 cm,f=−20 cmv2=\mathrm{B}, \mathrm{u}_2=-25 \mathrm{~cm}, \mathrm{f}=-20 \mathrm{~cm} \mathrm{v}_2= ?

amp;v2=u2fu2−f=(−25)(−20)(−25)−(−20)=500−5amp;v2=−100 cm(−ve,real)\begin{aligned}& \mathrm{v}_2=\frac{\mathrm{u}_2 \mathrm{f}}{\mathrm{u}_2-\mathrm{f}}=\frac{(-25)(-20)}{(-25)-(-20)}=\frac{500}{-5} \\\\& \mathrm{v}_2=-100 \mathrm{~cm}(- ve, real)\end{aligned}

Hence, the distance between images formed by the mirror is,

d = 60 – (–100) = 160 cm

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