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- The refractive index of the lens is 2.5
- The radius of curvature of the convex surface is 45 cm
- The faint image is erect and real
- The focal length of the lens is 20 cm

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Solution

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160+130=1f1

f1=20cm

Case II,

110−130=1f2

f2=15=R2

R=30 cm

From case I,

f1=Rn−1=20

Thus, n=2.5

Radius of curvature is 30 cm

Focal length of lens is f1=20 cm

Answer is option A and D.

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