Geometrical Optics — JEE Main practice

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Sample questions with solutions

Q1 · 2026

A convex lens is made from glass material having refractive index of 1.41.4 with same radius of curvature on both sides. The ratio of its focal length and radius of curvature is ____.

  • A.

    0.5

  • B.

    2.5

  • C.

    1.25

  • D.

    0.8

Answer: C
  1. For a thin lens, the Lens Maker's formula is
1f=(n−1)(1R1−1R2),\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right),

where:

  • ff is the focal length,
  • nn is the refractive index of the lens material,
  • R1R_1 and R2R_2 are the radii of curvature of the two surfaces.
  1. The lens is equi-convex, so both surfaces have the same radius of curvature RR.

According to the Cartesian sign convention,

R1=+R,R2=−R.R_1=+R,\qquad R_2=-R.
  1. Substituting these values,
1f=(1.4−1)(1R−1−R).\frac{1}{f} = (1.4-1) \left( \frac{1}{R} - \frac{1}{-R} \right).

Simplifying,

1f=0.4(2R)=0.8R.\frac{1}{f} = 0.4 \left( \frac{2}{R} \right) = \frac{0.8}{R}.
  1. Therefore,
f=R0.8.f=\frac{R}{0.8}.

Hence,

fR=10.8=1.25.\frac{f}{R} = \frac{1}{0.8} = 1.25.
  1. Therefore, the required ratio is 1.25, so the correct option is C.
Q2 · 2026

A prism of angle 75∘75^\circ and refractive index 3\sqrt{3} is coated with thin film of refractive index 1.51.5 only at the back exit surface. To have total internal reflection at the back exit surface the incident angle must be ______

(sin⁡15∘=0.25 and sin⁡25∘=0.43)\left(\sin15^\circ=0.25\ \text{and}\ \sin25^\circ=0.43\right)

  • A.

    between 15∘15^\circ and 20∘20^\circ

  • B.

    15∘15^\circ

  • C.

    <15∘<15^\circ

  • D.

    >25∘>25^\circ

Answer: A, B, C

  1. Total internal reflection (TIR) occurs at the interface between the prism and the thin film.

    The critical angle is given by

    sin⁡C=nrarerndenser=1.53=32.\sin C=\frac{n_{\text{rarer}}}{n_{\text{denser}}} =\frac{1.5}{\sqrt{3}} =\frac{\sqrt{3}}{2}.

    Therefore,

    C=60∘.C=60^\circ.

  2. For TIR at the coated surface, the angle of incidence inside the prism must satisfy

    r2>60∘.r_2>60^\circ.

  3. For a prism,

    A=r1+r2.A=r_1+r_2.

    Since

    A=75∘,A=75^\circ,

    we obtain

    r1=75∘−r2<15∘.r_1=75^\circ-r_2<15^\circ.

  4. Applying Snell's law at the first surface,

    nairsin⁡i=nprismsin⁡r1,n_{\text{air}}\sin i=n_{\text{prism}}\sin r_1,

    or

    sin⁡i=3sin⁡r1.\sin i=\sqrt{3}\sin r_1.

    As

    r1<15∘,r_1<15^\circ,

    we have

    sin⁡i<3sin⁡15∘=1.732×0.25≈0.433.\sin i<\sqrt{3}\sin15^\circ =1.732\times0.25 \approx0.433.

  5. Since

    sin⁡25∘=0.43,\sin25^\circ=0.43,

    it follows that

    i<25∘.i<25^\circ.

    Also, as r1r_1 can vary from 0∘0^\circ to just below 15∘15^\circ, the corresponding incident angle can vary from 0∘0^\circ to just below 25∘25^\circ.

  6. Therefore:

    • A. between 15∘15^\circ and 20∘20^\circ ✓
    • B. 15∘15^\circ ✓
    • C. <15∘<15^\circ ✓
    • D. >25∘>25^\circ ✗

    Hence, the correct options are A, B and C.

Q3 · 2026

As shown in the diagram, when the incident ray is parallel to the base of the prism, the emergent ray grazes along the second surface.

If the refractive index of the material of the prism is 2\sqrt{2}, the angle θ\theta of the prism is:

  • A.

    75∘75^\circ

  • B.

    90∘90^\circ

  • C.

    60∘60^\circ

  • D.

    45∘45^\circ

Answer: C
  1. The incident ray is parallel to the base of the prism. Since the base angle shown in the figure is
45∘,45^\circ,

the angle of incidence at the first face is also

i=45∘.i=45^\circ.

The refractive index of the prism is

μ=2.\mu=\sqrt{2}.
  1. Applying Snell's law at the first surface,
1⋅sin⁡45∘=2sin⁡r1.1\cdot\sin45^\circ = \sqrt{2}\sin r_1.

Substituting the value of sin⁡45∘\sin45^\circ,

12=2sin⁡r1,\frac{1}{\sqrt{2}} = \sqrt{2}\sin r_1,

which gives

sin⁡r1=12.\sin r_1=\frac{1}{2}.

Therefore,

r1=30∘.r_1=30^\circ.
  1. The emergent ray grazes the second surface, so the angle of emergence is
e=90∘.e=90^\circ.

Hence, the internal angle of incidence at the second face equals the critical angle.

Applying Snell's law,

2sin⁡r2=1⋅sin⁡90∘,\sqrt{2}\sin r_2 = 1\cdot\sin90^\circ, 2sin⁡r2=1,\sqrt{2}\sin r_2=1,

so

sin⁡r2=12,\sin r_2=\frac{1}{\sqrt{2}},

which gives

r2=45∘.r_2=45^\circ.
  1. For a prism,
A=r1+r2.A=r_1+r_2.

Therefore,

A=30∘+45∘=75∘.A=30^\circ+45^\circ=75^\circ.
  1. From the geometry of the prism,
45∘+θ+A=180∘.45^\circ+\theta+A=180^\circ.

Substituting A=75∘A=75^\circ,

45∘+θ+75∘=180∘,45^\circ+\theta+75^\circ=180^\circ, θ=180∘−120∘=60∘.\theta=180^\circ-120^\circ=60^\circ.

Therefore, the angle of the prism is

60∘.\boxed{60^\circ}.

Therefore, the correct answer is Option C.

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