Question

there is an equiconvex lens of focal length of 20cm. If the lens is cut into two equal parts perpendicular to the principle axis, the focal length of each part will be (a)20 cm (b)10 cm (c)40 cm (d)15 cm​

Answer 1

Answer:

C

Explanation:

lens formula :

1/f = (u-1)(1/R¹-1/R²)

u = refractive index

as equi convex R¹= R²

so 1/f = (u-1)(2/R)

when we cut it along perpendicular to principle axis radii would differ

R¹=R some constant but

R² = infinity as it is like a plane mirror

so now focal length becomes

1/f¹=(u-1)(1/R)

f¹=2f

Answer 2

Answer:

Explanation:

Solution,

Here, we have

Focal length of equiconvex lens = 20 cm

Means,

⇒ 1/f = (μ - 1) × (1/R - 1/R)

And for equiconvex lens,

⇒ R₁ = - R₂ = R

Then, here we have

⇒ 1/20 = (μ - 1) × (1/R - 1/R)

⇒ R = 40 (μ - 1)

Then the lens is cut into two equal parts perpendicular to the optical axis,

Then, here we have,

R₁ = ∞ and R₂ = R

Thus, we get

⇒ (1/f' = (μ - 1) × (1/R - 1/R)

⇒ 1/f' = (μ - 1) × (1/R)

Putting R's value, we get

⇒ 1/f' = (μ - 1) × [1/40(μ - 1)]

⇒ f' = 40 cm

Hence, the focal length of each part will be 40 cm.