Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Two transparent media A and B are separated by a plane boundary. The speeds of light in those media are $1.5 \times 10^8$ m/s and $2.0 \times 10^8$ m/s, respectively. The critical angle for a ray of light for these two media is:

tan⁻¹(0.750)

sin⁻¹(0.500)

sin⁻¹(0.750)

tan⁻¹(0.500)

Step-by-Step Solution

Refractive Index and Speed: The refractive index ( $n$ ) of a medium is inversely proportional to the speed of light ( $v$ ) in that medium ( $n = c/v$ ). Thus, the medium with the lower speed is optically denser.

Speed in A ( $v_A$ ) = $1.5 \times 10^8$ m/s (Denser)
Speed in B ( $v_B$ ) = $2.0 \times 10^8$ m/s (Rarer)

Condition for Total Internal Reflection: Light must travel from the denser medium (A) to the rarer medium (B).
Critical Angle Formula: The sine of the critical angle ( $i_c$ ) is equal to the ratio of the refractive indices ( $n_{rarer}/n_{denser}$ ) or the ratio of speeds ( $v_{denser}/v_{rarer}$ ). $\sin i_c = \frac{n_B}{n_A} = \frac{v_A}{v_B}$
Calculation: $\sin i_c = \frac{1.5 \times 10^8}{2.0 \times 10^8} = \frac{1.5}{2.0} = 0.75$ $i_c = \sin^{-1}(0.750)$

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