Question

an equiconvex lens has a power of 4 dioptre. what will be the radius of curvatures of each surface if the lens is made of flint glass of refractive index of 1.5?

Answer 1

Answer:-

• Power of lens=p=4D

Focal length=f:-

$\boxed{\sf\star P=\dfrac{1}{f}}$

$\\ \sf\longmapsto f=\dfrac{1}{p}$

$\\ \sf\longmapsto f=\dfrac{1}{4}$

$\\ \sf\longmapsto f=0.25m$

$\\ \sf\longmapsto f=25cm$

• Radius of curvature=R=?

$\boxed{\sf R=2f}$

$\\ \sf\longmapsto R=2(25)$

$\\ \sf\longmapsto R=50cm$

Answer 2

Given information to us:

• Equiconvex lens has a power of 4 dioptre.

Need to calculate:

• Radius of curvatures of each surface if the lens is made of flint glass of refractive index of 1.5.g ?

Solving way:

»» Here first we would be using the formula of calculating the power of lens that is,

Power of lens = 1 / focal length. Here after we would apply all the values in this formula. But we aren't asked to find out the power of lens as it is already given. After this we would reverse the sides in that formula thus focal length would be equal to 1/4.

»» Now after we would get the values of focal length we would calculate the radius of curvature by multiplying the 2 with focal length.

Calculations:

A thick lens which is of short focal length and the deviation of rays is more, whereas a thinner lens is of large focal length in this case deviation of rays is less.

Thus power of lens is represented in terms of reciprocal of focal length.

Unit of power of lens:

➟ Dioptre (D)

Here using the formula of calculating the power of lens that is as follows,

➟ Power of lens (D) = 1 / focal length

Substituting values :

➟ 4 = 1 / focal length

Acq,

➟ focal length = 1 / 4

Dividing :

4 ( 1 0 ) . 25

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎8

‎ ‎ ‎ ‎ ‎ ‎____

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎2 0

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎2 0

‎ ‎ ‎ ‎ ‎ ‎ ____

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎ ‎ 0

Now the focal length is in metres we will convert into centimetres :

➟ 0.25 × 100

➟ ( 25 / 100 ) × 100

➟ 25 cm

Thus the focal length in centimetres is 25 cm.

➟ 25 cm

Here finding out the radius of curvature:

➟ Radius of curvature = 2 × focal length

➟ Radius of curvature = 2 × 25 cm

➟ Radius of curvature = 50 cm

Thus,

• Radius of curvature is of 50 cm$\:$