The power of a thin convex lens ( a ng=1.5) is +5.0 D. When it is placed in a liquid of refractive index a ne, then it behaves as a concave lens of focal length 100 cm. The refractive index of the liquid a nl will be

Question

The power of a thin convex lens ( a ng=1.5) is +5.0 D. When it is placed in a liquid of refractive index a ne, then it behaves as a concave lens of focal length 100 cm. The refractive index of the liquid a nl will be (A) 5/3 (B) 4/3 (C) √3 (D) 5/4

Tardigrade Admin · Accepted Answer

We have, From lens maker formula, 5=(1.5-1)((2/R)) .....(i) and -1=((1.5/n)-1)((1/R)) ......(ii) Dividing Eq. (i) by Eq. (ii), we get -5=(0.5 n/1.5-n) Or -7.5+5 n=0.5 n Or -7.5=-4.5 n ∴ a ne=n=(75/45)=(5/3)

Q. The power of a thin convex lens $(_{a} n_{g} = 1.5)$ is $+ 5.0 D$ . When it is placed in a liquid of refractive index $_{a} n_{e}$ , then it behaves as a concave lens of focal length $100 c m$ . The refractive index of the liquid $_{a} n_{l}$ will be

5/3

4/3

$3$

5/4

We have, From lens maker formula, $5 = (1.5 - 1) (\frac{2}{R})$ .....(i)
and $- 1 = (\frac{1.5}{n} - 1) (\frac{1}{R})$ ......(ii)
Dividing Eq. (i) by Eq. (ii), we get
$- 5 = \frac{0.5 n}{1.5 - n}$
Or $- 7.5 + 5 n = 0.5 n$
Or $- 7.5 = - 4.5 n$
$∴_{a} n_{e} = n = \frac{75}{45} = \frac{5}{3}$

Q. The power of a thin convex lens (a​ng​=1.5) is +5.0D. When it is placed in a liquid of refractive index a​ne​, then it behaves as a concave lens of focal length 100cm. The refractive index of the liquid a​nl​ will be

5/3

4/3

3​

5/4

We have, From lens maker formula, 5=(1.5−1)(R2​) .....(i) and −1=(n1.5​−1)(R1​) ......(ii) Dividing Eq. (i) by Eq. (ii), we get −5=1.5−n0.5n​ Or −7.5+5n=0.5n Or −7.5=−4.5n ∴a​ne​=n=4575​=35​

Q. The power of a thin convex lens $(_{a} n_{g} = 1.5)$ is $+ 5.0 D$ . When it is placed in a liquid of refractive index $_{a} n_{e}$ , then it behaves as a concave lens of focal length $100 c m$ . The refractive index of the liquid $_{a} n_{l}$ will be

$3$

We have, From lens maker formula, $5 = (1.5 - 1) (\frac{2}{R})$ .....(i)
and $- 1 = (\frac{1.5}{n} - 1) (\frac{1}{R})$ ......(ii)
Dividing Eq. (i) by Eq. (ii), we get
$- 5 = \frac{0.5 n}{1.5 - n}$
Or $- 7.5 + 5 n = 0.5 n$
Or $- 7.5 = - 4.5 n$
$∴_{a} n_{e} = n = \frac{75}{45} = \frac{5}{3}$