An object AB is placed 15 text{ cm} on the le | vedclass

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(B) For lens $P$: $u_1 = -15 	ext{ cm}$, $f_1 = 10 	ext{ cm}$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$, we get $\frac

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real, formed at $7.5 	ext{ cm}$ right of lens $Q$, with a size same as that of $AB$

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(B) For lens $P$: $u_1 = -15 	ext{ cm}$, $f_1 = 10 	ext{ cm}$. Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$, we get $\frac{1}{v_1} - \frac{1}{-15} = \frac{1}{10} \implies \frac{1}{v_1} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30} \implies v_1 = 30 	ext{ cm}$.This image acts as a virtual object for lens $Q$. The distance between the lenses is $15 	ext{ cm}$. Since the image is formed $30 	ext{ cm}$ to the right of lens $P$, it is located $30 - 15 = 15 	ext{ cm}$ to the right of lens $Q$. Thus, $u_2 = +15 	ext{ cm}$.For lens $Q$: $u_2 = +15 	ext{ cm}$, $f_2 = 15 	ext{ cm}$. Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$, we get $\frac{1}{v_2} - \frac{1}{15} = \frac{1}{15} \implies \frac{1}{v_2} = \frac{2}{15} \implies v_2 = 7.5 	ext{ cm}$.Since $v_2 > 0$, the image is real and formed $7.5 	ext{ cm}$ to the right of lens $Q$.Total magnification $M = m_1 	imes m_2 = (v_1/u_1) 	imes (v_2/u_2) = (30/-15) 	imes (7.5/15) = -2 	imes 0.5 = -1$. The magnitude of magnification is $1$, so the size is the same as that of $AB$.

An object $AB$ is placed $15 \text{ cm}$ on the left of a convex lens $P$ of focal length $10 \text{ cm}$. Another convex lens $Q$ is now placed $15 \text{ cm}$ to the right of lens $P$. If the focal length of lens $Q$ is $15 \text{ cm}$, the final image is . . . . . . .

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Find the position of the final image with respect to the rightmost lens formed by the lens combination shown in the figure.

$A$ thin convex lens of focal length $30 \, cm$ forms an image $2 \, cm$ high,of an object at infinity. $A$ thin concave lens of focal length $20 \, cm$ is placed $26 \, cm$ from the convex lens on the side of the image. The height of the final image is.....$cm$.

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