A thin convex lens of focal length 5 cm and | vedclass

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Question

(A) For the combined lenses without a gap: The equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{5} - \f

Vedclass · Accepted Answer

$5/6$

Vedclass · Answer

(A) For the combined lenses without a gap: The equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{5} - \frac{1}{4} = -\frac{1}{20} \ cm$. So, $F = -20 \ cm$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$, with $u = -10 \ cm$, we get $\frac{1}{v} = \frac{1}{-20} + \frac{1}{-10} = -\frac{3}{20}$, so $v_1 = -\frac{20}{3} \ cm$. Magnification $m_1 = \frac{v_1}{u} = \frac{-20/3}{-10} = \frac{2}{3}$.For the lenses separated by $d = 1 \ cm$: First lens (convex) forms an image at $v'$ where $\frac{1}{v'} - \frac{1}{-10} = \frac{1}{5} \implies \frac{1}{v'} = \frac{1}{10} \implies v' = 10 \ cm$. Magnification $m_{1}' = \frac{v'}{u} = \frac{10}{-10} = -1$. This image acts as an object for the concave lens. The distance of this image from the concave lens is $u' = v' - d = 10 - 1 = 9 \ cm$. Since it is behind the lens, $u' = +9 \ cm$. Using the lens formula for the concave lens: $\frac{1}{v''} - \frac{1}{9} = \frac{1}{-4} \implies \frac{1}{v''} = \frac{1}{9} - \frac{1}{4} = -\frac{5}{36} \implies v'' = -\frac{36}{5} \ cm$. Magnification $m_{2}' = \frac{v''}{u'} = \frac{-36/5}{9} = -\frac{4}{5}$. Total magnification $m_2 = m_{1}' 	imes m_{2}' = (-1) 	imes (-4/5) = 4/5$. Finally, $\left|\frac{m_1}{m_2}ight| = \left|\frac{2/3}{4/5}ight| = \frac{2}{3} 	imes \frac{5}{4} = \frac{5}{6}$.

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