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 42

MEDIUM

12th Karnataka Board

IMPORTANT

Earn 100

Obtain the equation for apparent depth.

Important Questions on Ray Optics and Optical Instruments

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12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Ray Optics and Optical Instruments>Exercise>Q 43

What is optical density and how is it different from mass density?

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12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Ray Optics and Optical Instruments>Exercise>Q 44

A

4.5 c m

needle is placed

12 c m

away from a convex mirror of focal length

15 c m

. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

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12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Ray Optics and Optical Instruments>Exercise>Q 45

Use the mirror equation to deduce that:
an object placed between

f

and

2 f

of a concave mirror produces a real image beyond

2 f

.
[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

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12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Ray Optics and Optical Instruments>Exercise>Q 46

Answer the following questions:
You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.

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12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Ray Optics and Optical Instruments>Exercise>Q 47

Define the centre of curvature of a spherical mirror?

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12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 9 - Ray Optics and Optical Instruments>Exercise>Q 48

Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in the figure below. A current in the coil produces a deflection of $3.5 °$ of the mirror. What is the displacement of the reflected spot of light on a screen placed $1.5 m$ away?

Question Image