JEE Main 2014 · 12 Apr · Shift 1 · Physics Q23

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PhysicsSingle correct+4 / -1Question 23 of 90

The refractive index of the material of a concave lens is $\mu$ . It is immersed in a medium of refractive index $\mu_1$ . A parallel beam of light is incident on the lens. The path of the emergent rays when $\mu_1>\mu$ is :

The emergent rays converge to a point on the axis beyond the lens

A concave lens of index μ in a medium of index μ₁, with three parallel rays incident from the left; the emergent rays converge to a point on the axis to the right.

The emergent rays diverge away from the axis

The rays pass through undeviated

A concave lens of index μ in a medium of index μ₁, with three parallel rays incident from the left; the rays pass straight through undeviated.

The emergent rays bend towards and cross the axis, continuing beyond

A concave lens of index μ in a medium of index μ₁, with three parallel rays incident from the left; the upper ray bends down across the axis while the others continue undeviated.

Answer and worked solution

Official answer: A

Lens-maker's formula: $\dfrac1f=\left(\dfrac\mu{\mu_1}-1\right)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ . For a concave lens the second bracket is negative; with $\mu_1>\mu$ the first is negative too, so $f>0$ .

The lens converges: the emergent rays meet at a point on the axis, as in figure (1).

Topics:Thin lenses and lens maker's formula

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JEE Main 2014 · 12 Apr · Shift 1, Physics question 23