The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm . Its focal length is 12 cm . The refractive index of the glass is

Question

The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm . Its focal length is 12 cm . The refractive index of the glass is (A) 1.2 (B) 1.3 (C) 1.4 (D) 1.5

Tardigrade Admin · Accepted Answer

(1/f)=(μ-1)[(1/R1)-(1/R2)] ⇒ (μ-1)[(1/+10)-(1/(-15))] =(1/12)=(μ-1)((1/10)+(1/15))=(1/12) ⇒ μ=1.5

Q. The radii of curvature of the faces of a double convex lens are $10 c m$ and $15 c m .$ Its focal length is $12 c m .$ The refractive index of the glass is

1.2

1.3

1.4

1.5

$\frac{1}{f} = (μ - 1) [\frac{1}{R _{1}} - \frac{1}{R _{2}}]$
$\Rightarrow (μ - 1) [\frac{1}{+ 10} - \frac{1}{( - 15 )}]$
$= \frac{1}{12} = (μ - 1) (\frac{1}{10} + \frac{1}{15}) = \frac{1}{12}$
$\Rightarrow μ = 1.5$

Q. The radii of curvature of the faces of a double convex lens are 10cm and 15cm. Its focal length is 12cm. The refractive index of the glass is

1.2

1.3

1.4

1.5

f1​=(μ−1)[R1​1​−R2​1​] ⇒(μ−1)[+101​−(−15)1​] =121​=(μ−1)(101​+151​)=121​ ⇒μ=1.5

Q. The radii of curvature of the faces of a double convex lens are $10 c m$ and $15 c m .$ Its focal length is $12 c m .$ The refractive index of the glass is

$\frac{1}{f} = (μ - 1) [\frac{1}{R _{1}} - \frac{1}{R _{2}}]$
$\Rightarrow (μ - 1) [\frac{1}{+ 10} - \frac{1}{( - 15 )}]$
$= \frac{1}{12} = (μ - 1) (\frac{1}{10} + \frac{1}{15}) = \frac{1}{12}$
$\Rightarrow μ = 1.5$