Focal length of a thin convex lens is 30 cm. At a distance of 10 cm from the lens there is a plane refracting surface of refractive index 3/2. Where will the parallel rays incident on lens converge?

Question

Focal length of a thin convex lens is 30 cm. At a distance of 10 cm from the lens there is a plane refracting surface of refractive index 3/2. Where will the parallel rays incident on lens converge? (A) At a distance of 27.5 cm from the lens (B) At a distance of 25 cm from the lens (C) At a distance of 45 cm from the lens (D) At a distance of 40 cm from the lens

Tardigrade Admin · Accepted Answer

The lens will converge the rays at its focus i.e.,

30 c m

from the lens or

20 c m

from the refracting surface.

P I_{1} = 20 c m

∴ P I_{2} = μ (P I_{1})

= \frac{3}{2} \times 20 = 30 c m

So, the rays will converge at a distance of

40 c m

from the lens.

Q. Focal length of a thin convex lens is 30cm. At a distance of 10cm from the lens there is a plane refracting surface of refractive index 3/2. Where will the parallel rays incident on lens converge?

At a distance of 27.5 cm from the lens

At a distance of 25 cm from the lens

At a distance of 45 cm from the lens

At a distance of 40 cm from the lens

The lens will converge the rays at its focus i.e., 30cm from the lens or 20cm from the refracting surface. PI1​=20cm ∴PI2​=μ(PI1​) =23​×20=30cm So, the rays will converge at a distance of 40cm from the lens.

Q. Focal length of a thin convex lens is $30 c m$ . At a distance of $10 c m$ from the lens there is a plane refracting surface of refractive index $3/2$ . Where will the parallel rays incident on lens converge?

The lens will converge the rays at its focus i.e., $30 c m$ from the lens or $20 c m$ from the refracting surface.

$P I_{1} = 20 c m$
$∴ P I_{2} = μ (P I_{1})$
$= \frac{3}{2} \times 20 = 30 c m$
So, the rays will converge at a distance of $40 c m$ from the lens.