Three identical convex lenses each of focal l | vedclass

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(A) For the first lens:$u_1 = -f/2$, $f_1 = f$Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:$\frac{1}{v_1} - \frac{1}{-f/2} =

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Final image will be at $f/2$ behind the rightmost lens and its magnification will be $-1$.

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(A) For the first lens:$u_1 = -f/2$, $f_1 = f$Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$:$\frac{1}{v_1} - \frac{1}{-f/2} = \frac{1}{f} \Rightarrow \frac{1}{v_1} + \frac{2}{f} = \frac{1}{f} \Rightarrow \frac{1}{v_1} = -\frac{1}{f} \Rightarrow v_1 = -f$.Magnification $m_1 = \frac{v_1}{u_1} = \frac{-f}{-f/2} = 2$.For the second lens:The image from the first lens acts as an object for the second lens. The distance between the lenses is $f$. So, $u_2 = -(f + |v_1|) = -(f + f) = -2f$.Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$:$\frac{1}{v_2} - \frac{1}{-2f} = \frac{1}{f} \Rightarrow \frac{1}{v_2} + \frac{1}{2f} = \frac{1}{f} \Rightarrow \frac{1}{v_2} = \frac{1}{2f} \Rightarrow v_2 = 2f$.Magnification $m_2 = \frac{v_2}{u_2} = \frac{2f}{-2f} = -1$.For the third lens:The image from the second lens acts as an object for the third lens. The distance between the second and third lens is $f$. The image $v_2 = 2f$ is formed behind the second lens. The object distance for the third lens is $u_3 = -(f - v_2) = -(f - 2f) = f$.Using the lens formula $\frac{1}{v_3} - \frac{1}{u_3} = \frac{1}{f_3}$:$\frac{1}{v_3} - \frac{1}{f} = \frac{1}{f} \Rightarrow \frac{1}{v_3} = \frac{2}{f} \Rightarrow v_3 = f/2$.Magnification $m_3 = \frac{v_3}{u_3} = \frac{f/2}{f} = 1/2$.Total magnification $M = m_1 	imes m_2 	imes m_3 = 2 	imes (-1) 	imes (1/2) = -1$.The final image is at $f/2$ behind the rightmost lens.

$List-I$	$List-II$
$(P)$ If $n=2$ and $\alpha=180^{\circ}$,then all the possible values of $\theta_0$ will be	$(1)$ $30^{\circ}$ and $0^{\circ}$
$(Q)$ If $n=\sqrt{3}$ and $\alpha=180^{\circ}$,then all the possible values of $\theta_0$ will be	$(2)$ $60^{\circ}$ and $0^{\circ}$
$(R)$ If $n=\sqrt{3}$ and $\alpha=180^{\circ}$,then all the possible values of $\phi_0$ will be	$(3)$ $45^{\circ}$ and $0^{\circ}$
$(S)$ If $n=\sqrt{2}$ and $\theta_0=45^{\circ}$,then all the possible values of $\alpha$ will be	$(4)$ $150^{\circ}$
	$(5)$ $0^{\circ}$

Three identical convex lenses each of focal length $f$ are placed in a straight line separated by a distance $f$ from each other. An object is located at $f/2$ in front of the leftmost lens. Then,

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