Focal length of a convex lens of refraction index 1.5 is 2 cm. The foc

1.	Focal length of a convex lens of refraction index 1.5 is 2 cm. The focal length of lens, when immersed in a liquid of refractive index of 1.25, will be.(a) 10 cm(b) 2.5 cm(c) 5 c(d) 7.5 cm
Answer» Answer is (c) 5 c \(\frac{1}{f_{air}}=\Big(\frac{1.5}{1}-1\Big)\) \(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) ......(1) \(\frac{1}{f_{liquid}}=\Big(\frac{1.5}{1.25}-1\Big)\)\(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) ......(2) eq(2)/eq(1) \(\frac{f_{air}}{f_{liquid}}\) = \(\frac{2}{5}\) , \(f_{liquid}=\frac{5}{2}f_{air}=\frac{5}{2}\times2\) = 5cm

Focal length of a convex lens of refraction index 1.5 is 2 cm. The focal length of lens, when immersed in a liquid of refractive index of 1.25, will be.(a) 10 cm(b) 2.5 cm(c) 5 c(d) 7.5 cm

Answer»

Answer is (c) 5 c

\(\frac{1}{f_{air}}=\Big(\frac{1.5}{1}-1\Big)\) \(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) ......(1)

\(\frac{1}{f_{liquid}}=\Big(\frac{1.5}{1.25}-1\Big)\)\(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) ......(2)

eq(2)/eq(1)

\(\frac{f_{air}}{f_{liquid}}\) = \(\frac{2}{5}\) , \(f_{liquid}=\frac{5}{2}f_{air}=\frac{5}{2}\times2\) = 5cm