Find the time taken by light to cover a distance of 9 mm in water? Take $μ_{w a t e r}$ = $\frac{4}{3}$

0.04 ns

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0.4 ns

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4 ns

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400 ns

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Solution

The correct option is A 0.04 ns
Using snell's law,
$μ_{a i r} v_{a i r} = μ_{w a t e r} v$
$1 \times 3 \times 10^{8} = \frac{4}{3} \times v$
$v = \frac{9}{4} \times 10^{8} m / s$

From definition of speed,
$v = \frac{d}{t}$
$t = \frac{d}{v}$
$t = \frac{9 \times 10^{- 3} m}{\frac{9}{4} \times 10^{8} m / s}$
$t = 4 \times 10^{- 11} s$
$t = 0.04 n s$

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Similar questions

Find the time taken by light to cover a distance of 9 mm in water. Take $μ_{w a t e r}$ = $\frac{4}{3}$

Q. The ratio of the time taken by light to travel a distance of 100 m in air and water is

(Given

c = 3 \times 10^{8}

m/s ;

μ_{a i r} = 1.0003

;

μ_{w a t e r} = \frac{4}{3}

)

Q. The time taken by a monochromatic light ray to travel a certain distance in air is 9 x 10

^{- 6}

sec. The time taken by the light ray to travel the same distance in water of Refractive Index 4/3 is :

Q. A parallel narrow beam of light travelling through water in a container incident on an air bubble of radius

8

mm. What be the distance of image obtained due to refraction of beam on first face of air bubble from the centre of bubble? (take

μ_{w a t e r} = \frac{4}{3}

)

Question 35

An auto rickshaw takes 3 hours to cover a distance of 36 km. If its speed is increased by 4 km/h, the time taken by it to cover the same distance is___.

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Find the time taken by light to cover a distance of 9 mm in water? Take μwater = 43

The correct option is A 0.04 nsUsing snell's law, μairvair=μwaterv 1×3×108=43×v v=94×108 m/s From definition of speed, v=dt t=dv t=9× 10−3 m94× 108 m/s t=4×10−11 s t=0.04 ns

Find the time taken by light to cover a distance of 9 mm in water? Take $μ_{w a t e r}$ = $\frac{4}{3}$