Ray Optics

JEE Main level, numerical answer, medium

34 of 295

Question 34 of 295

An object 'OO' is placed at a distance of 100cm100cm in front of a concave mirror of radius of curvature 200cm200cm as shown in the figure. The object starts moving towards the mirror at a speed 2cms−12cm{s}^{-1}. The position of the image from the mirror after 10s10s will be at _____ cmcm.

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Answer: 400

In 10s10s, the object will travel distance =2×10=20cm=2\times 10=20cm
After 10sec⁡10\sec, the object will be at distance u=20−100=−80mu=20-100=-80m.
Focal length of the concave mirror is f=R2=2002=−100cmf=\dfrac{R}{2}=\dfrac{200}{2}=-100cm
Using mirror formula,
1v+1u=1f\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}
⇒1v=−1100+180⇒v=800020\Rightarrow \dfrac{1}{v}=-\dfrac{1}{100}+\dfrac{1}{80}\Rightarrow v=\dfrac{8000}{20}
⇒v=400cm\Rightarrow v=400cm
Position of image from the mirror is at 400cm400cm.

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