A biconvex lens of radius of curvature R is m | vedclass

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(A) The refractive index of the lens varies with distance $y$ from the principal axis as $\mu(y) = 2[1 + \frac{|y|}{d}]$.For a thin lens,the lens make

Vedclass · Accepted Answer

$6$

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(A) The refractive index of the lens varies with distance $y$ from the principal axis as $\mu(y) = 2[1 + \frac{|y|}{d}]$.For a thin lens,the lens maker's formula is $\frac{1}{f} = (\mu - 1)(\frac{2}{R})$.At the principal axis $(y = 0)$,$\mu = 2[1 + 0] = 2$. Thus,$\frac{1}{f_1} = (2 - 1)(\frac{2}{R}) = \frac{2}{R}$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,with $u = -R$,we get $\frac{1}{v_1} - \frac{1}{-R} = \frac{2}{R} \implies \frac{1}{v_1} = \frac{1}{R} \implies v_1 = R$.At the edge of the lens $(y = d)$,$\mu = 2[1 + \frac{d}{d}] = 4$. Thus,$\frac{1}{f_2} = (4 - 1)(\frac{2}{R}) = \frac{6}{R}$.Using the lens formula,$\frac{1}{v_2} - \frac{1}{-R} = \frac{6}{R} \implies \frac{1}{v_2} = \frac{5}{R} \implies v_2 = \frac{R}{5}$.The spread of the image is the difference in the image positions: $\Delta v = |v_1 - v_2| = |R - \frac{R}{5}| = \frac{4R}{5}$.Given the image spread is $6 \ m$,we have $\frac{4R}{5} = 6 \implies R = 7.5 \ m$.However,the question asks for the value of $n$ where the spread is $n$ meters. Since the spread is given as $6 \ m$,$n = 6$.

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