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CBSE Class 12 Physics 2020 Outside Delhi Set 1 Paper

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Question : 34 of 37
Marks: +1, -0
Two objects PP and QQ when placed at different positions in front of a concave mirror of focal length 20 cm20\,\mathrm{cm}, form real images of equal size. Size of object PP is three times the size of object QQ. If the distance of PP is 50 cm50\,\mathrm{cm} from the mirror, find the distance of QQ from the mirror.
Solution:  
Mirror formula : 1v+1u=1f\frac{1}{v}+\frac{1}{u}=\frac{1}{f}
Magnification=m=h′h=vu\text{Magnification}=m=\frac{h'}{h}=\frac{v}{u}
For object PP :
up=−50 cmu_{p}=-50\,\mathrm{cm}
f=−20 cmf=-20\,\mathrm{cm}
1vp+1up=1f\frac{1}{v_p}+\frac{1}{u_p}=\frac{1}{f}
Or, 1vp+1−50=1−20\text{Or, }\frac{1}{v_p}+\frac{1}{-50}=\frac{1}{-20}
∴vP=−1003 cm\therefore v_P=-\frac{100}{3}\,\mathrm{cm}
mP=hP′hP=vPuP=−1003−50=23m_{P}=\frac{h_P'}{h_P}=\frac{v_P}{u_P}=\frac{-\frac{100}{3}}{-50}=\frac{2}{3}
For object QQ :
mQ=hQ′hQ=vQuQm_{Q}=\frac{h_Q'}{h_Q}=\frac{v_Q}{u_Q}
Or, hp′13hP=vQuQ\frac{h_p'}{\frac{1}{3}h_P}=\frac{v_Q}{u_Q}
∴vQ=uQ\therefore v_Q=u_Q
1vQ+1uQ=1f\frac{1}{v_Q}+\frac{1}{u_Q}=\frac{1}{f}
12uQ+1uQ=1−20\frac{1}{2u_Q}+\frac{1}{u_Q}=\frac{1}{-20}
∴ uQ=−30 cmu_Q=-30\,\mathrm{cm}
So, the distance of QQ from mirror is 30 cm30\,\mathrm{cm} .
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