The image of an object placed in air formed b | vedclass

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(A) Given,the image distance $v = 10 \ m$ (behind the surface,so $v = +10 \ m$ by sign convention).The image is real,and the distance of the object $u

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) Given,the image distance $v = 10 \ m$ (behind the surface,so $v = +10 \ m$ by sign convention).The image is real,and the distance of the object $u$ is such that $v = \frac{2}{3} |u|$.Thus,$10 = \frac{2}{3} |u| \Rightarrow |u| = 15 \ m$. Since the object is in front of the surface,$u = -15 \ m$.The refractive index $\mu$ is given by the ratio of wavelengths: $\lambda_m = \frac{\lambda_a}{\mu} \Rightarrow \mu = \frac{\lambda_a}{\lambda_m} = \frac{1}{2/3} = 1.5 = \frac{3}{2}$.Using the refraction formula for a spherical surface: $\frac{\mu}{v} - \frac{1}{u} = \frac{\mu - 1}{R}$.Substituting the values: $\frac{3/2}{10} - \frac{1}{-15} = \frac{3/2 - 1}{R}$.$\frac{3}{20} + \frac{1}{15} = \frac{1/2}{R}$.$\frac{9 + 4}{60} = \frac{1}{2R} \Rightarrow \frac{13}{60} = \frac{1}{2R}$.$26R = 60 \Rightarrow R = \frac{60}{26} = \frac{30}{13} \ m$.Given $R = \frac{x}{13} \ m$,comparing the two,we get $x = 30$.

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