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 $\sin 45.0^\circ = 0.707$ )

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

Cevap

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

Adım Adım Çözüm

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.

Anahtar Kavram

Total Internal Reflection and Critical Angle

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru