A ray of light enters a rectangular glass slab of refractive index $\sqrt{3}$ at an angle of incidence $60^{O}$ . It travels a distance of 5 cm inside the slab and emerges out of the slab. The perpendicular distance between the incident and the emergent rays is

5 \sqrt{3}

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\frac{5}{2}

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5 \sqrt{\frac{3}{2}}

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5 cm

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Solution

The correct option is B $\frac{5}{2}$ cm
The perpendicular distance between the incident and the emergent ray is given by :

MN = t sec r sin (i – r)
and we know that, t=d cos r

Placing the known quantities in the formula we get the perpendicular distance as $\frac{5}{2} c m$

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A ray of light enters a rectangular glass slab of refractive index √3 at an angle of incidence 60O. It travels a distance of 5 cm inside the slab and emerges out of the slab. The perpendicular distance between the incident and the emergent rays is

The correct option is B 52cmThe perpendicular distance between the incident and the emergent ray is given by : MN = t sec r sin (i – r) and we know that, t=d cos r Placing the known quantities in the formula we get the perpendicular distance as 52cm

A ray of light enters a rectangular glass slab of refractive index $\sqrt{3}$ at an angle of incidence $60^{O}$ . It travels a distance of 5 cm inside the slab and emerges out of the slab. The perpendicular distance between the incident and the emergent rays is

The correct option is B $\frac{5}{2}$ cm
The perpendicular distance between the incident and the emergent ray is given by :

MN = t sec r sin (i – r)
and we know that, t=d cos r

Placing the known quantities in the formula we get the perpendicular distance as $\frac{5}{2} c m$