Question

A thin convex lens of refractive index $1.5$ has $20 \,cm$ focal length in air. If the lens is completely immersed in a liquid of refractive index $1.6$, its focal length will be

• A. - 160 cm
• B. - 100 cm
• C. #ERROR!
• D. #ERROR!

Answer 1

Answer: (A)

- 160 cm

Answer 2

From lens formula $\frac{1}{f} =\left(_{a} \mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ $=(1.5-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \dots$(i) Also, $_{l}\mu_{g}=\frac{\mu_{g}}{\mu_{l}}=\frac{1.5}{1.6}$ $\therefore \frac{1}{f'}=\left(_l\mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ $\frac{1}{f'}=\left(\frac{1.5}{1.6}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\dots$(ii) Dividing E (i) by E (ii), we get $\frac{f}{f'}=\frac{\left(\frac{1.5}{1.6}\right)}{(1.5-1)}=-\frac{1}{16 \times 0.5}$ $f'=-16 \times 0.5 \times f$ $=16 \times 0.5 \times 20$ $\Rightarrow f'=-160\, cm$