A convex lens of focal length 30,cm,a concave | vedclass

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(A) $1$. For the convex lens: $f_1 = +30\,cm$,$u_1 = -60\,cm$. Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we get $\frac{1}{30} =

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$60\,cm$ from the convex lens

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(A) $1$. For the convex lens: $f_1 = +30\,cm$,$u_1 = -60\,cm$. Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we get $\frac{1}{30} = \frac{1}{v_1} - \frac{1}{-60} \Rightarrow \frac{1}{v_1} = \frac{1}{30} - \frac{1}{60} = \frac{1}{60}$. Thus,$v_1 = +60\,cm$. This image acts as an object for the concave lens.$2$. The distance between the convex and concave lens is $20\,cm$. The image formed by the convex lens is $60\,cm$ behind it. Therefore,the distance of this image from the concave lens is $60 - 20 = 40\,cm$. Since it is behind the lens,it acts as a virtual object for the concave lens,so $u_2 = +40\,cm$.$3$. For the concave lens: $f_2 = -120\,cm$,$u_2 = +40\,cm$. Using the lens formula: $\frac{1}{-120} = \frac{1}{v_2} - \frac{1}{40} \Rightarrow \frac{1}{v_2} = \frac{1}{40} - \frac{1}{120} = \frac{3-1}{120} = \frac{2}{120} = \frac{1}{60}$. Thus,$v_2 = +60\,cm$. This image is formed $60\,cm$ behind the concave lens.$4$. The plane mirror is at a distance of $70\,cm$ from the convex lens. Since the concave lens is $20\,cm$ from the convex lens,the mirror is $70 - 20 = 50\,cm$ from the concave lens. The image $v_2$ is $60\,cm$ behind the concave lens,which is $60 - 50 = 10\,cm$ behind the mirror. This acts as a virtual object for the plane mirror.$5$. The plane mirror forms a real image at $10\,cm$ in front of it. Since the mirror is $50\,cm$ from the concave lens,the final image is $50 - 10 = 40\,cm$ behind the concave lens,or $60\,cm$ behind the convex lens.

$A$ convex lens of focal length $30\,cm$,a concave lens of focal length $120\,cm$,and a plane mirror are arranged as shown. For an object kept at a distance of $60\,cm$ from the convex lens,the final image,formed by the combination,is a real image at a distance of:

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