26. Focal length of a convex lens of refractive index 1.5 is 2 cm. Focal length of the lens when immersed in a liquid of refractive index of 1.25 will be ?
(a) 10 cm
(b) 2.5 cm
(c) 5 cm
(d) 7.5 cm
Answer : (c) 5 cm
\(\frac{1}{f} = (n_2 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\)
\(\frac{1}{f} = (n_{21} - 1) \frac{2}{R} \quad \text{equiconvex}\)
\(n_1 = 1.5,\; n_2 = 1.5,\; n_3 = 1.25\)
\(n_{21} = \frac{n_2}{n_1}\)
\(n_{23} = \frac{n_2}{n_3}\)
\(\frac{1}{f} = (n_{21} - 1) \frac{2}{R}\)
\(\frac{1}{2} = (1.5 - 1) \frac{2}{R}\)
\(\Rightarrow R = 2\)
\(\frac{1}{f_{\text{liq}}} = (n_{23} - 1) \frac{2}{R}\)
\(\frac{1}{f_{\text{liq}}} = \left( \frac{1.5}{1.25} - 1 \right) \frac{2}{R}\)
\(\Rightarrow f_{\text{liq}} = 5\)
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