Question

# A point-like object is placed at a distance of $$1 m$$ in front of a convex lens of focal length $$0.5 m$$. A plane mirror is placed at a distance of $$2 m$$ behind the lens. The position and nature of the final image formed by the system is

A
$$2.6 m$$ from the mirror, real
B
$$1 m$$ from the mirror, real
C
$$2.6 m$$ from the mirror, virtual
D
$$1 m$$ from the mirror, virtual
Solution
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#### Correct option is A. $$2.6 m$$ from the mirror, realFor first reflection at convex lens$$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$$$$\Rightarrow \dfrac{1}{+0.5} = \dfrac{1}{v} - \dfrac{1}{-1}$$$$\Rightarrow \dfrac{1}{v} = 2- 1$$$$\Rightarrow c = -1m$$.Real image is formed at $$1m$$ from the lens.This image acts as an object for the plane mirror after reflection, it's image is formed at $$1m$$ behind the plane mirror at $$I'$$The image at $$I'$$ is the virtual object for the convex lens.So, $$\mu = -3 m$$ for $$I'$$$$f = +0.5m$$ for $$I'$$For second refraction at convex lens.$$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$$$$\Rightarrow \dfrac{1}{0.5} = \dfrac{1}{v} - \dfrac{1}{-3}$$$$\Rightarrow 2 - \dfrac{1}{3} = \dfrac{1}{v}$$$$\Rightarrow \dfrac{5}{3} = \dfrac{1}{v}$$$$\Rightarrow v = + \dfrac{3}{5}$$For $$v = \dfrac{3}{5} = 0.6$$ (towards left of lens).So distance from mirror $$\Rightarrow 2 + 0.6 = 2.6m$$ (real image) 3
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