A converging mirror is placed on the right-ha | vedclass

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(C) For the final beam to be parallel to the principal axis after passing through the lens, the light rays incident on the lens from the mirror must a

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) For the final beam to be parallel to the principal axis after passing through the lens, the light rays incident on the lens from the mirror must appear to come from the focal point of the lens. Thus, the image formed by the mirror must be at the focal point of the lens, which is $15 \ cm$ to the left of the lens.Since the mirror is $40 \ cm$ to the right of the lens, the image formed by the mirror is at a distance $v_2 = -(40 - 15) = -25 \ cm$ from the mirror (using sign convention where light travels left to right, and the mirror is on the right).For the mirror, the focal length $f_m = -20 \ cm$ (concave mirror).Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:$\frac{1}{-25} + \frac{1}{u_2} = \frac{1}{-20} \Rightarrow \frac{1}{u_2} = \frac{1}{25} - \frac{1}{20} = \frac{4-5}{100} = -\frac{1}{100}$.So, $u_2 = -100 \ cm$. This means the object for the mirror is $100 \ cm$ to the right of the mirror.The image formed by the lens $(v_1)$ acts as the object for the mirror. Since the mirror is $40 \ cm$ from the lens, the object distance for the mirror is $u_2 = v_1 - 40$. Thus, $v_1 - 40 = -100 \Rightarrow v_1 = -60 \ cm$.Now, use the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_l}$:$\frac{1}{-60} - \frac{1}{-d} = \frac{1}{15} \Rightarrow \frac{1}{d} = \frac{1}{15} + \frac{1}{60} = \frac{4+1}{60} = \frac{5}{60} = \frac{1}{12}$.Therefore, $d = 12 \ cm$.

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