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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

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For 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)

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