The size of the image of an object at infinit | vedclass

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(B) Given: Focal length of convex lens $f_{1} = 30 \,cm$, focal length of concave lens $f_{2} = -20 \,cm$, and initial image size $h_{o} = 2 \,cm$.Sin

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$2.5$

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(B) Given: Focal length of convex lens $f_{1} = 30 \,cm$, focal length of concave lens $f_{2} = -20 \,cm$, and initial image size $h_{o} = 2 \,cm$.Since the object is at infinity, the image formed by the convex lens is at its focus, $v_{1} = 30 \,cm$.The concave lens is placed at a distance of $26 \,cm$ from the convex lens. Thus, the image formed by the convex lens acts as a virtual object for the concave lens.The distance of this virtual object from the concave lens is $u_{2} = v_{1} - 26 = 30 - 26 = 4 \,cm$.Using the lens formula for the concave lens: $\frac{1}{v_{2}} - \frac{1}{u_{2}} = \frac{1}{f_{2}}$.Substituting the values: $\frac{1}{v_{2}} - \frac{1}{4} = \frac{1}{-20}$.$\frac{1}{v_{2}} = \frac{1}{4} - \frac{1}{20} = \frac{5-1}{20} = \frac{4}{20} = \frac{1}{5}$.So, $v_{2} = 5 \,cm$.The magnification $m$ produced by the concave lens is $m = \frac{v_{2}}{u_{2}} = \frac{5}{4} = 1.25$.The new size of the image $h_{i} = m 	imes h_{o} = 1.25 	imes 2 \,cm = 2.5 \,cm$.

The size of the image of an object at infinity, formed by a convex lens of focal length $30 \,cm$, is $2 \,cm$. If a concave lens of focal length $20 \,cm$ is placed between the convex lens and the image at a distance of $26 \,cm$ from the convex lens, what is the new size of the image (in $\,cm$)?

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