Question

A plano convex lens of focal length
16cm is to be made of glass of refraction
Index 1.5 The radius of curvature
of the curved surface is​

Answer 1

Given :

• Focal length of plano - convex lens = 16 cm
• Refractive index of lens = 1.5

To Find :

The radius of curvature of the curved surface

Theory :

Lens Maker's Formula :

$\sf\green{\dfrac{1}{f}=(_{m}u_{l}-1)[\dfrac{1}{R_1}-\dfrac{1}{R_2}]}$

Where , $\sf_{m}u_{l}$ refractive index of lens with respect to medium .

• $\sf\dfrac{1}{R_1}$ is Radius of curvature of first surface of lens.
• $\sf\dfrac{1}{R_2}$ is Radius of curvature of second surface of lens.

Solution:

We have ,

• $\sf_{m}u_{l}=1.5$
• $\sf\:R_1=?$
• $\sf\:R_2=\infty$
• Focal length of lens , f = 16

We have to find Radius of curvature of curved surface i.e, $\sf\:R_1$

By Lens - Maker's Formula

$\sf\blue{\dfrac{1}{f}=(_{m}u_{l}-1)[\dfrac{1}{R_1}-\dfrac{1}{R_2}]}$

Now , Put the given values

$\sf\implies\dfrac{1}{16}=(1.5-1)[\dfrac{1}{R_1}-\dfrac{1}{\infty}]$

$\sf\implies\dfrac{1}{16}=(0.5)[\dfrac{1}{R_1}-0]$

$\sf\implies\dfrac{1}{16}=(0.5)\times\dfrac{1}{R_1}$

$\sf\implies\:R_1=0.5\times16$

$\sf\implies\:R_1=\dfrac{5}{10}\times16$

$\sf\implies\:R_1=\dfrac{1}{2}\times16$

$\sf\implies\:R_1=8cm$

Therefore , The radius of curvature of the curved surface is 8 cm

Answer 2

${\large{\bold{\sf{\underline{Understanding \; the \; question}}}}}$

➝ This question says that a plane convex lens of focal length 16 cm is to be made of glass of refraction. Index 1.5 We have to find the radius of curvature of the curved surface.

${\large{\bold{\sf{\underline{Given \; that}}}}}$

➝ Focal length of plane convex lens is 16 cm

➝ Index is 1.5

${\large{\bold{\sf{\underline{To \; find}}}}}$

➝ The radius of curvature of the curved surface.

${\large{\bold{\sf{\underline{Solution}}}}}$

➝ The radius of curvature of the curved surface = 8 cm

${\large{\bold{\sf{\underline{Using \: concept}}}}}$

➝ Len's maker formula

${\large{\bold{\sf{\underline{Using \: formula}}}}}$

${\bold{\sf{\longmapsto Len's \: maker = \dfrac{1}{f} = (_{m}u_{I}-1) [\dfrac{1}{R_1} - \dfrac{1}{R_2}}}}$

${\large{\bold{\sf{\underline{These \; denotes}}}}}$

➝ ${\sf_{m}u_{l}}$ means refractive index of lens, respect to the medium.

➝ ${\sf{\dfrac{1}{R_1}}}$ means radius of curvature of the first surface of lens.

➝ ${\sf{\dfrac{1}{R_2}}}$ means radius of curvature of the second surface of lens.

➝ f means focal length

${\large{\bold{\sf{\underline{Solution}}}}}$

~ Here we have –

➝ ${\sf_{m}u_{l}}$ as 1.5

➝ ${\sf{\dfrac{1}{R_1}}}$ as ?

➝ ${\sf{\dfrac{1}{R_2}}}$ as ∞

➝ Focal length of lens as 16 cm

~ Finding ${\sf{\dfrac{1}{R_1}}}$

${\bold{\sf{\longmapsto Len's \: maker = \dfrac{1}{f} = (_{m}u_{I}-1) [\dfrac{1}{R_1} - \dfrac{1}{R_2}}}}$

${\bold{\sf{\longmapsto Len's \: maker = \dfrac{1}{16} = (1.5 - 1) [\dfrac{1}{R_1} - \dfrac{1}{\infin}]}}}$

${\bold{\sf{\longmapsto Len's \: maker = \dfrac{1}{16} = (0.5) [\dfrac{1}{R_1} - 0]}}}$

${\bold{\sf{\longmapsto Len's \: maker = \dfrac{1}{16} = (0.5) \times \dfrac{1}{R_1}}}}$

${\bold{\sf{\longmapsto R_{1} = 0.5 \times 16}}}$

${\bold{\sf{\longmapsto R_{1} = 8 \: cm}}}$