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(D) The light undergoes three phenomena:$(i)$ Refraction from the lens.$(ii)$ Reflection from the mirror.$(iii)$ Refraction from the lens.$1^{	ext{st

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None of these

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(D) The light undergoes three phenomena:$(i)$ Refraction from the lens.$(ii)$ Reflection from the mirror.$(iii)$ Refraction from the lens.$1^{	ext{st}}$ Refraction from the lens:Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,where $u = -40\, cm$ and $f = +20\, cm$:$\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} \Rightarrow v = +40\, cm$.The magnification $m_1 = \frac{v}{u} = \frac{40}{-40} = -1$.$2^{	ext{nd}}$ Reflection from the mirror:The image $I_1$ acts as an object for the mirror. The distance of $I_1$ from the mirror is $60\, cm - 40\, cm = 20\, cm$. So,$u = -20\, cm$ and $f = -10\, cm$ (for a concave mirror).Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v} = \frac{1}{-10} - \frac{1}{-20} = -\frac{1}{20} \Rightarrow v = -20\, cm$.The magnification $m_2 = -\frac{v}{u} = -\frac{-20}{-20} = -1$.$3^{	ext{rd}}$ Refraction from the lens:The image $I_2$ acts as an object for the lens. The distance of $I_2$ from the lens is $60\, cm - 20\, cm = 40\, cm$. So,$u = -40\, cm$ and $f = +20\, cm$.Using the lens formula:$\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} \Rightarrow v = +40\, cm$.The magnification $m_3 = \frac{v}{u} = \frac{40}{-40} = -1$.Total magnification $m = m_1 	imes m_2 	imes m_3 = (-1) 	imes (-1) 	imes (-1) = -1$.The final image is formed at a distance of $40\, cm$ from the convergent lens,which is at the same position as the original object,and it has the same size as the object. Since none of the options match this result,the correct answer is $D$.

An upright object is placed at a distance of $40\, cm$ in front of a convergent lens of focal length $20\, cm$. $A$ convergent mirror of focal length $10\, cm$ is placed at a distance of $60\, cm$ on the other side of the lens. The position and size of the final image will be

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