#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# A ray of light moving along the vector (-i^-2j^) undergoes refraction at an interface of two media, which is the x-z plane. The refractive index for y>0 is 2 while for y<0 its [root(5)]/2. The unit vector along which the reflected light ray moves is ?

2 years ago

A sketch will show you the vector (-1)î + (-2)ĵ makes an angle θ₁ to the normal of the interface where
tan(θ₁) = (-1)/(-2) = 1/2
θ₁ = tan⁻¹(1/2)

If you sketch a right angled triangle with perpendicular sides of lengths 1 and 2 you will see the hypotenuse is length √(1² +2²) = √5. It follows that sin(θ₁) = 1/√5

If the angle of refraction is θ₂ we can apply Snell's Law:
n₁sin(θ₁) =n₂sin(θ₂)
2 X 1/√5 =√5/2 x sin(θ₂)
2 = 5/2 x sin(θ₂)
θ₂ = sin⁻¹(4/5).

We recognise that in a '3, 4, 5 triangle θ₂ = sin⁻¹(4/5),.means that tan(θ₂) = 4/3.

From our initial refraction diagram, this means the refracted ray must lie along the vector (-4)î + (-3)ĵ.

Since √(4² +3²) = 5, the unit vector along which the refracted light ray moves is (1/5)[(-4)î + (-