Three glass cylinders of equal height H = 30 | vedclass

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(B) For refraction at a single spherical surface,the formula is: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$. Here,$n_1 = 1.5$,$n_2 = 1$,and

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$1, 4$

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(B) For refraction at a single spherical surface,the formula is: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$. Here,$n_1 = 1.5$,$n_2 = 1$,and $u = -H = -30 	ext{ cm} = -0.3 	ext{ m}$.For cylinder $I$ (flat top): $R = \infty$. The apparent depth $H_1 = \frac{H}{n} = \frac{30}{1.5} = 20 	ext{ cm} = 0.2 	ext{ m}$.For cylinder $II$ (convex top): $R = +3 	ext{ m}$.$\frac{1}{v} - \frac{1.5}{-0.3} = \frac{1 - 1.5}{3} \implies \frac{1}{v} + 5 = -\frac{0.5}{3} = -\frac{1}{6} \implies \frac{1}{v} = -\frac{1}{6} - 5 = -\frac{31}{6} \implies v = -\frac{6}{31} 	ext{ m} \approx -19.35 	ext{ cm}$.Thus,$H_2 = 19.35 	ext{ cm}$.For cylinder $III$ (concave top): $R = -3 	ext{ m}$.$\frac{1}{v} - \frac{1.5}{-0.3} = \frac{1 - 1.5}{-3} \implies \frac{1}{v} + 5 = \frac{-0.5}{-3} = \frac{1}{6} \implies \frac{1}{v} = \frac{1}{6} - 5 = -\frac{29}{6} \implies v = -\frac{6}{29} 	ext{ m} \approx -20.69 	ext{ cm}$.Thus,$H_3 = 20.69 	ext{ cm}$.Comparing the values: $H_3 (20.69 	ext{ cm}) > H_1 (20 	ext{ cm}) > H_2 (19.35 	ext{ cm})$.Therefore,statements $(1)$ and $(4)$ are correct.

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