The radius of the curvature of the convex surface of a plano-convex lens is $20$ cm. If the refractive index of the material of the lens is $1.5$ . Its focal length is:

Question

The radius of the curvature of the convex surface of a plano-convex lens is 20 cm- If the refractive index of the material of the lens is 1-5- Its focal length is-10 cm20 cm30 cm40 cm

Toppr Admin · Accepted Answer

From Lens makers formula-xA0-1f-x3BC-x2212-1-1R1-x2212-1R2-R1-x221E-R2-x2212-20-xA0-cm1f-1-5-x2212-1-120-f-40cm

The radius of the curvature of the convex surface of a plano-convex lens is $20$ cm. If the refractive index of the material of the lens is $1.5$ . Its focal length is:
$10 c m$
$20 c m$
$30 c m$
$40 c m$

From Lens makers formula, $\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$R_{1} = \infty$

$R_{2} = - 20 c m$

$\frac{1}{f} = (1.5 - 1) (\frac{1}{20})$

$f = 40 c m$

The radius of the curvature of the convex surface of a plano-convex lens is 20 cm. If the refractive index of the material of the lens is 1.5. Its focal length is:10cm20cm30cm40cm

From Lens makers formula, 1f=(μ−1)(1R1−1R2)R1=∞R2=−20 cm1f=(1.5−1)(120)f=40cm

The radius of the curvature of the convex surface of a plano-convex lens is $20$ cm. If the refractive index of the material of the lens is $1.5$ . Its focal length is:
$10 c m$
$20 c m$
$30 c m$
$40 c m$

From Lens makers formula, $\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$R_{1} = \infty$

$R_{2} = - 20 c m$

$\frac{1}{f} = (1.5 - 1) (\frac{1}{20})$

$f = 40 c m$