Total internal reflection — find critical angle for glass (n=1.5) to water (n=1.33)

easy CBSE JEE-MAIN NEET • 3 min read
Tags Optics

Question

Find the critical angle for total internal reflection at a glass-water interface, given the refractive index of glass is 1.5 and of water is 1.33.

Solution — Step by Step

Total Internal Reflection (TIR) occurs when:

  1. Light travels from a denser medium to a less dense medium (higher nn to lower nn)
  2. The angle of incidence exceeds the critical angle θc\theta_c

Here, glass (n=1.5n = 1.5) is denser than water (n=1.33n = 1.33), so TIR is possible at this interface when light travels from glass into water.

At the critical angle, the refracted ray just grazes the interface (angle of refraction = 90°).

Applying Snell’s law:

n1sin⁡θc=n2sin⁡90°n_1 \sin\theta_c = n_2 \sin 90° n1sin⁡θc=n2×1n_1 \sin\theta_c = n_2 \times 1 sin⁡θc=n2n1\sin\theta_c = \frac{n_2}{n_1}

Here, n1=1.5n_1 = 1.5 (glass, denser), n2=1.33n_2 = 1.33 (water, less dense):

sin⁡θc=1.331.5=0.8867\sin\theta_c = \frac{1.33}{1.5} = 0.8867 θc=sin⁡−1(0.8867)\theta_c = \sin^{-1}(0.8867) θc≈62.5°\theta_c \approx 62.5°

So the critical angle at a glass-water interface is approximately 62.5°.

Any ray hitting the glass-water boundary from inside the glass at an angle greater than 62.5° will undergo total internal reflection. At angles less than 62.5°, some light is refracted into water (and some is reflected).

For comparison, the critical angle for glass-to-air (n_air = 1) is: sin⁡θc=1/1.5=0.667\sin\theta_c = 1/1.5 = 0.667 → θc≈41.8°\theta_c \approx 41.8° — much smaller.

Why This Works

Snell’s law (n1sin⁡θ1=n2sin⁡θ2n_1 \sin\theta_1 = n_2 \sin\theta_2) governs all refraction. At the critical angle, the refracted ray runs parallel to the interface (θ2=90°\theta_2 = 90°). Beyond this angle, there’s no refracted ray — all the light bounces back into the denser medium (total reflection). This is why TIR is “total” — unlike ordinary reflection, which always lets some light through.

The glass-water critical angle is larger than the glass-air critical angle because water and glass are closer in refractive index — the “bending” needed to exceed 90° refraction requires a larger incident angle.

JEE Main frequently tests: find critical angle given refractive indices, or identify which interface allows TIR. Remember — TIR is only possible from denser to less dense medium. Light in water hitting a glass surface from below cannot undergo TIR at the glass interface (water → glass is denser → denser, refraction always occurs).

Alternative Method

If you’re given the critical angle θc\theta_c and one refractive index, and asked to find the other:

n2=n1sin⁡θcn_2 = n_1 \sin\theta_c

This reverse calculation is common in JEE problems.

Common Mistake

Students sometimes apply sin⁡θc=n1/n2\sin\theta_c = n_1/n_2 instead of n2/n1n_2/n_1. The correct formula is sin⁡θc=n2/n1\sin\theta_c = n_2/n_1 where n1n_1 is the denser medium (glass) and n2n_2 is the less dense one (water/air). The smaller refractive index goes on top. If you swap them, you get sin⁡θc>1\sin\theta_c > 1, which is impossible — that’s your signal you’ve inverted the ratio.

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