A real image of half the size is obtained in a concave spherical mirror with a radius of curvature of 40 cm. The distance of the object and that of its image from the mirror will be respectively

15 cm, 30 cm

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30 cm, 15 cm

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30 cm, 60 cm

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60 cm, 30 cm

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Solution

The correct option is D 60 cm, 30 cm
Given, image formed is real.
$m = \frac{- v}{u} = - \frac{1}{2} \Rightarrow - \frac{u}{v} = \frac{1}{2} \Rightarrow v = \frac{u}{2}$
$\Rightarrow \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \Rightarrow \frac{1}{u} + \frac{2}{u} = - \frac{1}{20} [∵ f = \frac{R}{2}]$
$\Rightarrow \frac{1 + 2}{u} = - \frac{1}{20} \Rightarrow u = - 60 c m$
$∴ v = - 30 c m$ where v is the image distance, u is the object distance and m is the magnification.

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A real image of half the size is obtained in a concave spherical mirror with a radius of curvature of 40 cm. The distance of the object and that of its image from the mirror will be respectively

The correct option is D 60 cm, 30 cmGiven, image formed is real. m=−vu=−12⇒−uv=12⇒v=u2 ⇒1u+1v=1f⇒1u+2u=−120 [∵f=R2] ⇒1+2u=−120⇒u=−60 cm ∴v=−30 cm where v is the image distance, u is the object distance and m is the magnification.