A ray of light is incident parallel to BC at a height h=3.0 cm from BC. Find the height (in cm) above BC at which the emergent ray leaves the surface AC. If is given that $μ = \sqrt{2}$ and length BC=20 cm. Take $\tan 15^{0} = 0.25$ .

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\sin 45^{0} = μ \sin r \Rightarrow r = 30^{0}$

From $Δ B Q R$ and $Δ C S R$

$\frac{B Q}{S C} = \frac{B R}{R C} ∴ h' = 3 (\frac{R C}{B R}) i = 45^{0} + r = 75^{0}$

So, $\frac{Q M}{M R} = \tan 15^{0}$ , so MR = 12 cm
So, BR = 15 cm and RC = 5 cm
So h’ = 1 cm

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test