A glass rod has ends as shown in the figure. | vedclass

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(A) For the $1^{	ext{st}}$ refraction at the convex surface (radius $R$): Using $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$,where $

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$2 < \mu < 2.25$

Vedclass · Answer

(A) For the $1^{ ext{st}}$ refraction at the convex surface (radius $R$): Using $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$,where $\mu_1 = 1$,$\mu_2 = \mu$,$u = -2R$,and $R_1 = +R$. $\frac{\mu}{v_1} - \frac{1}{-2R} = \frac{\mu - 1}{R} \implies \frac{\mu}{v_1} = \frac{\mu - 1}{R} - \frac{1}{2R} = \frac{2\mu - 3}{2R} \implies v_1 = \frac{2\mu R}{2\mu - 3}$. For the $2^{ ext{nd}}$ refraction at the concave surface (radius $R/2$): The object distance $u_2 = v_1 - 3R = \frac{2\mu R}{2\mu - 3} - 3R = \frac{2\mu R - 6\mu R + 9R}{2\mu - 3} = \frac{R(9 - 4\mu)}{2\mu - 3}$. Here $\mu_1 = \mu$,$\mu_2 = 1$,and $R_2 = +R/2$ (center is to the right). Using $\frac{1}{v_f} - \frac{\mu}{u_2} = \frac{1 - \mu}{R/2} = \frac{2(1 - \mu)}{R}$. $\frac{1}{v_f} = \frac{2(1 - \mu)}{R} + \frac{\mu(2\mu - 3)}{R(9 - 4\mu)} = \frac{2(1 - \mu)(9 - 4\mu) + 2\mu^2 - 3\mu}{R(9 - 4\mu)} = \frac{18 - 8\mu - 18\mu + 8\mu^2 + 2\mu^2 - 3\mu}{R(9 - 4\mu)} = \frac{10\mu^2 - 29\mu + 18}{R(9 - 4\mu)}$. For a real image,$v_f > 0$. Since $R > 0$,we need $\frac{10\mu^2 - 29\mu + 18}{9 - 4\mu} > 0$. Factoring the numerator: $10\mu^2 - 20\mu - 9\mu + 18 = 10\mu(\mu - 2) - 9(\mu - 2) = (10\mu - 9)(\mu - 2)$. So,$\frac{(10\mu - 9)(\mu - 2)}{-(4\mu - 9)} > 0 \implies \frac{(10\mu - 9)(\mu - 2)}{4\mu - 9} The critical points are $\mu = 0.9, 2, 2.25$. Testing intervals,the inequality holds for $2 < \mu < 2.25$.

$A$ glass rod has ends as shown in the figure. The refractive index of glass is $\mu$. The object $O$ is at a distance $2R$ from the surface of larger radius of curvature. The distance between the apexes of the ends is $3R$. The range of $\mu$ for which the image is real is given by

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