Focal length of a convex lens of refractive index 1.5 is 2 cm. Focal length of lens when immersed in a liquid of refractive index of 1.25 will be

Question

Focal length of a convex lens of refractive index 1.5 is 2 cm. 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 cm (D) 7.5 cm

Tardigrade Admin · Accepted Answer

(fa/fe) = (( (μg/μl) -1)/(μg -1) ) = ( ((1.5/1.25) -1 )/1.5-1) = ((1/5)/(1)2) = (2/5) fe = (5/2) fa = (5/2) × 2 = 5 cm

Q. Focal length of a convex lens of refractive index 1.5 is 2 cm. Focal length of lens when immersed in a liquid of refractive index of 1.25 will be

10 cm

2.5 cm

5 cm

7.5 cm

$\frac{f _{a}}{f _{e}} = \frac{( \frac{μ _{g}}{μ _{l}} - 1 )}{( μ _{g} - 1 )} = \frac{( \frac{1.5}{1.25} - 1 )}{1.5 - 1} = \frac{\frac{1}{5}}{\frac{1}{2}} = \frac{2}{5}$
$f_{e} = \frac{5}{2} f_{a} = \frac{5}{2} \times 2 = 5 c m$

Q. Focal length of a convex lens of refractive index 1.5 is 2 cm. Focal length of lens when immersed in a liquid of refractive index of 1.25 will be

10 cm

2.5 cm

5 cm

7.5 cm

fe​fa​​=(μg​−1)(μl​μg​​−1)​=1.5−1(1.251.5​−1)​=21​51​​=52​ fe​=25​fa​=25​×2=5cm

$\frac{f _{a}}{f _{e}} = \frac{( \frac{μ _{g}}{μ _{l}} - 1 )}{( μ _{g} - 1 )} = \frac{( \frac{1.5}{1.25} - 1 )}{1.5 - 1} = \frac{\frac{1}{5}}{\frac{1}{2}} = \frac{2}{5}$
$f_{e} = \frac{5}{2} f_{a} = \frac{5}{2} \times 2 = 5 c m$