An object AB is placed 15 cm on the left of a convex lens P of focal length 10 cm

Question

An object AB is placed 15 cm on the left of a convex lens P of focal length 10 cm. Another convex lens Q is now placed 15 cm right of lens P. If the focal length of lens Q is 15 cm, the final image is ______. [2026]

Smart Achievers · Accepted Answer

(2)

$For lens P : u = - 15 cm$

$v_{1} = \frac{u f}{u + f} = \frac{- 15 \times 10}{- 15 + 10} = \frac{- 150}{- 5} = + 30$

$m_{1} = \frac{30}{- 15} = - 2$

$For lens Q : u = + 15 cm$

$v_{2} = \frac{u f}{u + f} = \frac{15 \times 15}{15 + 15} = \frac{15 \times 15}{30} = 7.5 cm$

$m_{2} = \frac{7.5}{15} = \frac{1}{2}$

$m = m_{1} m_{2} = - 2 \times \frac{1}{2} = - 1$

Solution

Q.
An object AB is placed 15 cm on the left of a convex lens P of focal length 10 cm. Another convex lens Q is now placed 15 cm right of lens P. If the focal length of lens Q is 15 cm, the final image is ______. [2026]

1 virtual, formed at 7.5 cm right of lens Q, with a size bigger than that of AB

2 real, formed at 7.5 cm right of lens Q, with a size same than that of AB

3 formed at infinity

4 real, formed at 7 cm right of lens Q, with a size smaller than that of AB

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