A ray of light traveling within a dense glass prism of refractive index 1.60 strikes the b… — JAMB UTME

A ray of light traveling within a dense glass prism of refractive index $1.60$ strikes the boundary with a surrounding transparent liquid. If total internal reflection just occurs at an angle of incidence of $45.0^\circ$ in the glass, what is the refractive index of the liquid? (Take $\sin 45.0^\circ = 0.707$ )

$1.13$ Answer
B
$2.26$
C
$0.44$
D
$1.60$

Answer

The refractive index of the liquid is

1.13

For light traveling from a denser medium (

n_{\text{glass}}

) to a rarer medium (

n_{\text{liquid}}

), the critical angle

\theta_c

is defined by

\sin \theta_c = \frac{n_{\text{liquid}}}{n_{\text{glass}}}

. Substituting

n_{\text{glass}} = 1.60

and

\sin 45.0^\circ = 0.707

gives

n_{\text{liquid}} = 1.60 \times 0.707 = 1.13

Step-by-Step Solution

Identify the given parameters and formula for total internal reflection

Refractive index of denser medium

n_{\text{glass}} = 1.60

, critical angle

\theta_c = 45.0^\circ

, and formula

\sin \theta_c = \frac{n_{\text{rarer}}}{n_{\text{denser}}}

Total internal reflection occurs at the critical angle when light travels from an optically denser medium to a less dense (rarer) medium.

Rearrange the equation to solve for the refractive index of the liquid (

n_{\text{liquid}}

)

n_{\text{liquid}} = n_{\text{glass}} \times \sin \theta_c

Multiplying both sides of the critical angle equation by

n_{\text{glass}}

isolates the target variable.

Substitute the values and calculate

n_{\text{liquid}}

n_{\text{liquid}} = 1.60 \times 0.707 = 1.1312 \approx 1.13

Carrying out the arithmetic yields the refractive index of the liquid.

Key Concept

Total Internal Reflection and Critical Angle

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