SATHEE: Optics 4 Question 7

Optics 4 Question 7

7. A right angled prism of refractive index $μ_{1}$ is placed in a rectangular block of refractive index $μ_{2}$ , which is surrounded by a medium of refractive index $μ_{3}$ , as shown in the figure, $A$ ray of light ’ $e$ ’ enters the rectangular block at normal incidence. Depending upon the relationships between $μ_{1}, μ_{2}$ and $μ_{3}$ , it takes one of the four possible paths ’ $e f$ ‘, ’ $e g$ ‘, ’ $e h$ ’ or ’ $e i$ ‘.

(2013 Adv.)

Match the paths in Column I with conditions of refractive indices in Column II and select the correct answer using the codes given below the lists.

Column I		Column II
P.	$e \to f$	1.	$μ_{1} > \sqrt{2} μ_{2}$
Q.	$e \to g$	2.	$μ_{2} > μ_{1}$ and $μ_{2} > μ_{3}$
R.	$e \to h$	3.	$μ_{1} = μ_{2}$
S.	$e \to i$	4.	$μ_{2} < μ_{1} < \sqrt{2} μ_{2}$ and $μ_{2} > μ_{3}$

Codes

P Q R S

(a) $2 3 1 4$

(b) $1 2 4 3$

(d) $2 3 4 1$

Show Answer

Answer:

Correct Answer: 7. (d)

Solution:

For $e \to i$

$\begin{array}{lrl} \Rightarrow & 45^{\circ} & > θ_{c} \\ \Rightarrow & \sin 45^{\circ} & > \sin θ_{c} \\ \Rightarrow & \frac{1}{\sqrt{2}} & > \frac{μ_{2}}{μ_{1}} \\ \Rightarrow & μ_{1} & > \sqrt{2} μ_{2} \end{array}$

For $e \to f$

angle of refraction is lesser than angle of incidence, so $μ_{2} > μ_{1}$ and then $μ_{2} > μ_{3}$

$\begin{aligned} For e \to g, μ_{1} = μ_{2} \\ for e \to h, μ_{2} < μ_{1} < \sqrt{2} μ_{2} and μ_{2} > μ_{3} \end{aligned}$

Optics 4 Question 7

Codes

Answer:

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Optics 4 Question 7

Codes

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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