Physics, asked by anashkjan123, 9 months ago

RAY OPTICS
Q.1 An object lies on the principal axis of a concave mirror with radius of curvature 160 cm. Its
vertical image appears at a distance 70 cm from it. Determine the position of the object and
also the magnification.​

Answers

Answered by Anonymous
5

Answer:

Given

Radius of curvature, R = 160 cm

Focal length, f = R/2 = 80 cm

Distance of the image, v = 70cm (sign is positive as virtual image)

By applying mirror's formula

\mathsf{\implies\frac{1}{f} = \frac{1}{v} + \frac{1}{u}}

\mathsf{\implies\frac{1}{u} = \frac{1}{f} - \frac{1}{v}}

\mathsf{\implies\frac{1}{u} = \frac{1}{80} - (\frac{1}{70})}

\mathsf{\implies\frac{1}{u} = \frac{70 - 80}{5600}}

\mathsf{\implies\frac{1}{u} = \frac{-10}{5600}}

\mathsf{\implies\frac{1}{u} = \frac{-1}{560}}

\fbox{\mathsf{ u = -560 cm}}

magnification = -v/u

\mathsf{ \implies\:m =  \frac{- (70) }{- 560}}

\mathsf{ \implies\: m =  \frac{1}{8}}

\fbox{\mathsf{ m =  + 0.125 }}

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