An object is placed at a distance of 5 , cm f | vedclass

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(D) The focal length of the convex lens is $f = +10 \, cm$. The first principal focus $(F_1)$ is at a distance of $10 \, cm$ from the optical center o

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$30$

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(D) The focal length of the convex lens is $f = +10 \, cm$. The first principal focus $(F_1)$ is at a distance of $10 \, cm$ from the optical center on the left side. The object is placed $5 \, cm$ from $F_1$ towards the lens,so the object distance $u = -(10 + 5) = -15 \, cm$. Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$. Substituting the values: $\frac{1}{v} - \frac{1}{-15} = \frac{1}{10}$. $\frac{1}{v} + \frac{1}{15} = \frac{1}{10}$. $\frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30}$. Therefore,$v = +30 \, cm$. The image is formed at a distance of $30 \, cm$ from the lens.

An object is placed at a distance of $5 \, cm$ from the first principal focus of a convex lens of focal length $10 \, cm$. If a real image is formed,find its distance from the lens in $cm$.

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