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Total Internal Reflection — Critical angle, optical fibre

Ray Optics

12

JEE Qs

8%

Hard

60

min

Thoroughly understand the conditions for Total Internal Reflection and practice problems involving critical angle calculations at various interfaces and in practical applications like optical fibers and prisms.

🧮 Key Formulas

n1 sin(i) = n2 sin(r) (Snell's Law)
sin(i_c) = n_rarer / n_denser
sin(i_c) = n2 / n1 (where n1 is refractive index of denser medium and n2 is refractive index of rarer medium)

✅ Key Points for JEE

  • 1Total Internal Reflection (TIR) occurs only when light travels from a denser medium to a rarer medium.
  • 2For TIR to occur, the angle of incidence (i) in the denser medium must be greater than the critical angle (i_c).
  • 3The critical angle (i_c) is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90 degrees.
  • 4The critical angle depends on the refractive indices of the two media at their interface; a smaller refractive index ratio (n_rarer/n_denser) means a smaller critical angle and thus a higher chance of TIR.
  • 5Optical fibers work on the principle of multiple total internal reflections, allowing light signals to travel long distances with minimal loss.

⚠️ Common Mistakes

  • Applying the condition for TIR when light is traveling from a rarer medium to a denser medium.
  • Incorrectly identifying n1 and n2 in the critical angle formula (n_rarer / n_denser vs. n_denser / n_rarer).
  • Confusing the angle of incidence with the angle from the surface, rather than with the normal.

📝 Practice Questions

See all

Q46.The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature R = 2 m . Another car approaches him from behind with a uniform speed of 90 km/hr. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the car in the side view mirror is ' a '. The value of 100 a is _______ m/s2 .

2025·NumericalHard

Q27.A spherical surface of radius of curvature R, separates air from glass (refractive index = 1.5 ). The centre of curvature is in the glass medium. A point object ' O ' placed in air on the optic axis of the surface, so that its real image is formed at ' I ' inside glass. The line OI intersects the spherical surface at P and PO = PI. The distance PO equals to (1) 5 R (2) 3 R (3) 1.5 R (4) 2 R

2025·MCQMedium

Q32.Given is a thin convex lens of glass (refractive index μ ) and each side having radius of curvature R. One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that 2025 (22 Jan Shift 1) JEE Main Previous Year Paper the image gets formed on the object itself ? (1) R/μ (2) R/(2μ −3) (3) μR (4) R/(2μ −1)

2025·MCQHard

Q42.In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to |R1| and |R2|, i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is (1) 1 − + |R2| |R2| 6 ( |R1|1 1 ) (2) −16 ( |R1|1 1 ) (3) 1 + − |R2| |R2| 6 ( |R1|1 1 ) (4) −16 ( |R1|1 1 )

2025·MCQMedium

Q29.A symmetric thin biconvex lens is cut into four equal parts by two planes AB and CD as shown in figure. If the power of original lens is 4 D then the power of a part of the divided lens is (1) D (2) 8D (3) 2D (4) 4D

2025·MCQMedium

Q29.Given a thin convex lens (refractive index μ2 ), kept in a liquid (refractive index μ1, μ1 < μ2 ) having radii of curvatures |R1| and |R2|. Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place? (1) μ1|R1|⋅|R2| (2) μ1|R1|⋅|R2| μ2(|R1|+|R2|)−μ1|R2| μ2(|R1|+|R2|)−μ1|R1| (3) (μ2+μ1)|R1| (4) μ1|R1|⋅|R2| (μ2−μ1) μ2(2|R1|+|R2|)−μ1√|R1|⋅|R2|

2025·MCQHard

NCERT Chapters

  • Class 12 Physics Ch 9: Ray Optics and Optical Instruments