Question

the power of an equiconvex lens with radii of curvature 5 cm and refractive index 1.5 in air is​

Answer 1

Explanation:

Power=p=

f

1

=(n−1)(

R

1

−

−R

1

)=0.6(2/R)=1.2/R=1.2/.1=+12 Diopters.

Answer 2

Answer:

The power of the given lens is 20 D.

Explanation:

Here we have been given that the radius of curvature of the given equiconvex lens is 5 cm and its refractive index is given to be 1.5 in air.

For the equiconvex lens, we have,

$R_{1} = R$

$R_{2} = - R$

Therefore the power for this lens can be calculated using the formula as below;

P = $\frac{1}{f}$ = (n-1) $(\frac{1}{R_{1} } - \frac{1}{R_{2} } )$

here n is the refractive index of the given lens and $R_{1}$ and $R_{2}$ are the radius of curvatures respectively of the lens surfaces.

GIven, n = 1.5 cm

$R_{1} = R = 5cm$

= 0.05 m

$R_{2} = - R = - 5cm$

= - 0.05m

∴ P = (1.5 - 1) $(\frac{1}{0.05 } - \frac{1}{-0.05 } )$

⇒ P = (0.5) × $(\frac{2}{0.05} )$

⇒ P = (10) × 2

⇒ P =  20 Dioptres

The power of the given equiconvex lens is found to be 20 D.