Skip to content
Question of 58

Q.A ray of light is incident at the glass-water interface at an angle of incidence i, if it emerges finally parallel to the surface of water, then the refractive index of glass μg would be

(a) (4/3) sin i
(b) 1/sin i
(c) 4/3
(d) 1
Meghalaya MboseMBOSE Meghalaya 11th Board 2019MCQ· 1mImportance★★★★★
0% · 0/58 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Applying Snell's law successively at the glass-water interface and then the water-air (top) surface — where the ray must refract at exactly 90 degrees to travel parallel to the surface — the intermediate water refractive index cancels, leaving glass index = 1/sin i. Note: this question is on Ray Optics, a topic not present in the 14-chapter menu supplied for this course (Ray Optics is a Class 12 NCERT chapter), so it is flagged out of syllabus for this course's chapter list, even though the physics itself is answerable.

Let n_g = refractive index of glass, n_w = refractive index of water, n_a = 1 (refractive index of air).

Step 1 — glass to water interface: the ray hits at angle of incidence i (in glass) and refracts into water at angle r.

By Snell's law: n_g sin i = n_w sin r ... (1)

Step 2 — this refracted ray then reaches the top (water-air) surface at angle of incidence r, and "emerges finally parallel to the surface of water" means it refracts into the air at 90 degrees (grazing along the surface).

By Snell's law at this second interface: n_w sin r = n_a sin 90 degrees = 1 x 1 = 1 ... (2)

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.