If the critical angle of the medium is 30-circ- the velocity of light in that medium is -velocity of light in air 3 times 10-8 m-s-6 times 10-8-m- s3 times 10-8-m -s1-5 times 10-8-m- s1 times 10-8-m-s

Question

If the critical angle of the medium is 30-circ- the velocity of light in that medium is -velocity of light in air 3 times 10-8  m-s-6 times 10-8-m- s3 times 10-8-m -s1-5 times 10-8-m- s1 times 10-8-m-s

Toppr Admin · Accepted Answer

As- -x3B8-cric-sin-x2212-1-1-x3BC-sin-x3B8-cric-1-x3BC-x3BC-1sin30-2Also as -x3BC-cvm-3-xD7-108vmvm-1-5-xD7-108m-s-

If the critical angle of the medium is $30^{\circ}$ , the velocity of light in that medium is :
(velocity of light in air $3 \times 10^{8}$ m/s)
$6 \times 10^{8} m / s$
$3 \times 10^{8} m / s$
$1.5 \times 10^{8} m / s$
$1 \times 10^{8} m / s$

As, $θ_{c r i c} = s i n^{- 1} (\frac{1}{μ})$

$s i n θ_{c r i c} = \frac{1}{μ}$

$μ = \frac{1}{s i n 30} = 2$

Also as $μ = \frac{c}{v_{m}} = \frac{3 \times 10^{8}}{v_{m}}$

$v_{m} = 1.5 \times 10^{8} m / s .$

If the critical angle of the medium is 30∘, the velocity of light in that medium is :(velocity of light in air 3×108 m/s)6×108m/s3×108m/s1.5×108m/s1×108m/s

As, θcric=sin−1(1μ)sinθcric=1μμ=1sin30=2Also as μ=cvm=3×108vmvm=1.5×108m/s.

If the critical angle of the medium is $30^{\circ}$ , the velocity of light in that medium is :
(velocity of light in air $3 \times 10^{8}$ m/s)
$6 \times 10^{8} m / s$
$3 \times 10^{8} m / s$
$1.5 \times 10^{8} m / s$
$1 \times 10^{8} m / s$

As, $θ_{c r i c} = s i n^{- 1} (\frac{1}{μ})$

$s i n θ_{c r i c} = \frac{1}{μ}$

$μ = \frac{1}{s i n 30} = 2$

Also as $μ = \frac{c}{v_{m}} = \frac{3 \times 10^{8}}{v_{m}}$

$v_{m} = 1.5 \times 10^{8} m / s .$