A biconvex lens of focal length 15cm is in front of a plane mirror. The distance between the lens and the mirror is 10cm. A small object is kept at a distance of 30 cm from the lens. The final image is

Engineering

Physics

Curved Surface Refraction

Question

real and at a distance of 16 cm from the mirror

virtual and at a distance of 16 cm from the mirror

virtual and at a distance of 20 cm from the mirror

real and at a distance of 20 cm from the mirror

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u = + 10

$\frac{1}{v} - \frac{1}{10} = \frac{1}{15}$

$\frac{1}{v} = \frac{4}{30} + \frac{3}{30}$

v = 6 cm