A thin convex lens of crown glass having refractive index 1.5 has power 1 D. What will be the power of similar convex lens of refractive index 1.6 ?

Question

A thin convex lens of crown glass having refractive index 1.5 has power 1 D. What will be the power of similar convex lens of refractive index 1.6 ? (A) 0.6 D (B) 0.8 D (C) 1.2 D (D) 1.6 D

Tardigrade Admin · Accepted Answer

P=(1/f)(n-1)((1/R1)-(1/R2)) Hence (P2/P1)=((n2-1)/(n1-1))That is (p2/1)=(1.6-1/1.5-1) Hence p2=1.2

Q. A thin convex lens of crown glass having refractive index 1.5 has power 1 D. What will be the power of similar convex lens of refractive index 1.6 ?

0.6 D

0.8 D

1.2 D

1.6 D

$P = \frac{1}{f} (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$He n ce \frac{P _{2}}{P _{1}} = \frac{( n _{2} - 1 )}{( n _{1} - 1 )} T ha t i s \frac{p _{2}}{1} = \frac{1.6 - 1}{1.5 - 1}$
$He n ce p_{2} = 1.2$

Q. A thin convex lens of crown glass having refractive index 1.5 has power 1 D. What will be the power of similar convex lens of refractive index 1.6 ?

0.6 D

0.8 D

1.2 D

1.6 D

P=f1​(n−1)(R1​1​−R2​1​) HenceP1​P2​​=(n1​−1)(n2​−1)​Thatis1p2​​=1.5−11.6−1​ Hencep2​=1.2

$P = \frac{1}{f} (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$
$He n ce \frac{P _{2}}{P _{1}} = \frac{( n _{2} - 1 )}{( n _{1} - 1 )} T ha t i s \frac{p _{2}}{1} = \frac{1.6 - 1}{1.5 - 1}$
$He n ce p_{2} = 1.2$