The figure shows a ray of light falling normally on the face AB of an equilateral glass prism having a refractive index 3 / 2 , placed in water of refractive index 4 / 3 . Will this ray suffer total internal reflection on striking the face AC? Justify your answer.

It is given that the glass prism is equilateral. μ p = 3 / 2 μ w = 4 / 3 Thus, ∠ A = 60 ° . Let us draw a normal to the face AC. From the geometry, we can see, the angle of incidence on the face AC is 60 ° . Thus, i = 60 ° . Let this angle be the critical angle i.e. i = i c . Then the refractive index of the prism in the air, μ = 1 sin i c = 1 sin 60 ° μ = 2 3 = 1 . 54 Now if the prism is placed in water then the refractive index of a prism with water, μ A = μ p μ w = 3 2 4 3 = 9 8 = 1 . 125 The total internal reflection at face AC, is possible only when μ < μ A . But from the results obtained above, we can see that μ > μ A ∴ 1 . 54 > 1 . 125 No total internal reflection takes place and the light ray gets refracted from the prism placed in water. Thus, we can conclude that for total internal reflection, in this case, the Critical angle should be greater than 60 ° .

An object is placed in front of a concave mirror. It is observed that a virtual image is formed. Draw the ray diagram to show the image formation and hence derive the mirror equation 1 f = 1 u + 1 v .

Ray diagram to show the formation of a virtual image by concave mirror: Derivation of mirror equation using the above ray diagram: ∆ C A ' B ' and ∆ C A B are similar triangles. Hence, we can write: A ' B ' A B = C B ' C B = C P + P B ' C P - B P .......(i) D N is equal to A B and is a normal on the principal axis. ∆ F D N and ∆ F A ' B ' are similar triangles. The incident rays are considered to be paraxial rays hence, D and P are very close. We can write: A ' B ' A B = F B ' F D = F P + P B ' F D = F P + P B ' F P .....(ii) Using the above equations and substituting the values according to the sign convention we get: C P + P B ' C P - B P = F P + P B ' F P - 2 f + v - 2 f + u = - f + v - f ⇒ - 2 f 2 - f v = 2 f 2 - f u - 2 v f = u v ⇒ f u + f v = u v f On dividing both sides by u v f we get: 1 u + 1 v = 1 f The above equation is the required mirror equation.

E M B I B E

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Ray Optics and Optical Instruments>Exercise>Q 80

HARD

12th Karnataka Board

IMPORTANT

Earn 100

What are non paraxial rays?

Important Questions on Ray Optics and Optical Instruments

MEDIUM

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Ray Optics and Optical Instruments>Exercise>Q 1

A container contains water up to a height of $20 cm$ and there is a point source at the centre of the bottom of the container. A rubber ring of radius $r$ floats centrally on the water. The ceiling of the room is $2.0 m$ above the water surface.

(a) Find the radius of the shadow of the radius of the ring formed on the ceiling if $r = 15 cm$ .

(b) Find the maximum value of $r$ for which the shadow of the ring is formed on the ceiling. Refractive index of water $= \frac{4}{3}$ .