Question

28. A lens made of glass of refractive index 1.5 has focal length 25 cm. Its focal length in water will be (refractive index of water is 4/3)
1
(1) 25 cm
(3) 68 cm
(2) 35 cm
(4) 100 cm​

Answer 1

Given:

Refractive index of glass (μ1) = 1.5

Focal length in glass (f1) = 25cm

Refractive index of water (μ2) = 4/3

Focal length in water (f2) =?

Solution:

Focal length in glass is given as,

1/f1=(μ1-1) (1/R1-1/R2) ---eq.(i)

where R1 and R2 are radii of curvatures of two surfaces of the lens

Focal length in water is given as,

1/f2=(μ2-1) (1/R1-1/R2) ----eq.(ii)

Dividing eq. (i) by eq. (ii)

f2/f1 = μ1-1/μ2-1

f2/25 = (1.5-1)/(4/3-1)

f2/25=0.5/0.33

f2=37.87cm

Hence, the focal length of lens in water will be 37.87cm.

Answer 2

Answer:

The focal length in water is 100cm.

Explanation:

Given the refractive index of a glass lens, $n_g = 1.5$

Focal length, $f_a = 25 cm$

Refractive index of water, $n_w = 4/3$

Focal length by lens maker's formula is given by

$\dfrac{1}{f} =(n-1)\Big(\dfrac{1}{R_1}-\dfrac{1}{R_2} \Big)$

where $R_1 \& R_2$ are radii of curvatures of the two surfaces of lens, and n is the refractive of the material.

When two different media are given, the refractive index of medium 1 with respect to the medium 1 is given by relative refractive index.    

And is written as $_2n_{1} =\dfrac{n_{1}}{n_{2}}$

Since the glass lens is kept in the water, we need to consider the relative refractive index.

Thus, when the lens is placed in air, the focal length is

$\dfrac{1}{f_a} =(_an_g-1)\Big(\dfrac{1}{R_1}-\dfrac{1}{R_2} \Big)\implies\dfrac{1}{f_a} =\Big(\dfrac{n_g}{n_a} -1\Big)\Big(\dfrac{1}{R_1}-\dfrac{1}{R_2} \Big)$

Similarly, when the lens is placed in water, the focal length is

$\dfrac{1}{f_w} =(_wn_g-1)\Big(\dfrac{1}{R_1}-\dfrac{1}{R_2} \Big)\implies\dfrac{1}{f_w} =\Big(\dfrac{n_g}{n_w} -1\Big)\Big(\dfrac{1}{R_1}-\dfrac{1}{R_2} \Big)$

As the lens is same, therefore $R_1 \& R_2$ are constant, and refractive index of sir, $n_a=1$. Thus,

$\dfrac{f_w}{f_a} =\frac{\Big(\dfrac{n_g}{n_a} -1\Big)}{\Big(\dfrac{n_g}{n_w} -1\Big)}$

Substituting the given values,

$\dfrac{f_w}{25} =\frac{\Big(\dfrac{1.5}{1} -1\Big)}{\Big(\dfrac{1.5}{4/3} -1\Big)}$

$f_w =\frac{({1.5} -1)\times 25}{\Big(\dfrac{1.5}{4/3} -1\Big)}$

$=\frac{({1.5} -1)\times 25}{\big({\frac{1.5\times3}{4} }-1\big)} =\frac{{0.5}\times 25}{\frac{0.5}{4}} =100m$

Therefore, the focal length in water is 100cm.