An object is situated at $8 c m$ from a convex lens of focal length $10 c m$ . Find the position and the nature of image. Draw ray diagram to illustrate the formation of the image

-40

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-20

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Solution

The correct option is A -40
$F r o m l e n s f o r m u l a e w e g e t \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \frac{1}{f} + \frac{1}{u} = \frac{1}{v} N o w o b j e c t d i s t a n c e f r o m o p t i c c e n t r e = - 8 c m a n d f o c a l l e n g t h = 10 c m \frac{1}{v} = \frac{1}{10} - \frac{1}{8} v = - 40 c m S i n c e - v e s i g n t h e i m a g e w i l l f o r m i n f r o n t o f t h e l e n s a n d w i l l v i r t u a l, e n l a r g e a n d e r e c t .$

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An object is situated at 8cm from a convex lens of focal length 10cm. Find the position and the nature of image. Draw ray diagram to illustrate the formation of the image

The correct option is A -40Fromlensformulaeweget1f=1v−1u1f+1u=1vNowobjectdistancefromopticcentre=−8cmandfocallength=10cm1v=110−18v=−40cmSince−vesigntheimagewillforminfrontofthelensandwillvirtual,enlargeanderect.

An object is situated at $8 c m$ from a convex lens of focal length $10 c m$ . Find the position and the nature of image. Draw ray diagram to illustrate the formation of the image