Animated Solution for Physics - Optics: A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of 60∘ (see figure). If the refractive index of the material of the prism is 3, which of the following is (are) correct?
Select Answer:
* Multiple Correct
Visualized Solution
\text{Analyzing the Setup}
\text{Prism angles: } \angle A = 90^{\circ}, \angle B = 60^{\circ}, \angle C = 135^{\circ}, \angle D = 75^{\circ}
\text{Incident ray } OP \text{ makes } 60^{\circ} \text{ with the normal at face } AB.
\text{Ray turns by } 180^{\circ} - 2(45^{\circ}) = 90^{\circ}
\text{New direction is } -120^{\circ} \text{ to the horizontal.}
\text{Angle of incidence at } AD \text{ is } i_3 = 30^{\circ}
\text{Emergence from Face } AD
\sqrt{3} \cdot \sin 30^{\circ} = 1 \cdot \sin e
\sin e = \frac{\sqrt{3}}{2} \implies e = 60^{\circ}
\text{Angle Between Incident and Emergent Rays}
\text{Incident ray direction: } -60^{\circ}
\text{Emergent ray direction: } -150^{\circ}
\text{Angle between them } = |-150^{\circ} - (-60^{\circ})| = 90^{\circ}
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The Sigma Insight: Refraction and Total Internal Reflection
Solution Diagram
This problem is a beautiful symphony of geometry and optics. It tests your ability to track a light ray through multiple interfaces, applying Snell's Law and checking for Total Internal Reflection (TIR) at every step. Let's embark on this optical journey!
Analyzing the Setup
We are given a quadrilateral prism ABCD with specific angles: ∠A=90∘, ∠B=60∘, ∠C=135∘, and ∠D=75∘. Let's orient the prism on a coordinate plane. If we place vertex A at the origin and align the base AD along the horizontal x-axis, the 90∘ angle at A means the face AB is perfectly vertical.
A monochromatic ray OP strikes this vertical face AB near vertex B. The problem states the angle of incidence is 60∘. Since the face AB is vertical, its normal is horizontal. Thus, the incident ray makes a 60∘ angle with the horizontal.
Refraction at Face AB
As the ray enters the prism, it bends towards the normal. We apply Snell's Law at the air-glass interface:
1⋅sin60∘=3⋅sinr1
Substituting sin60∘=23, we get:
23=3⋅sinr1⟹sinr1=21
This gives an angle of refraction r1=30∘. The refracted ray now travels inside the prism, making a 30∘ angle with the horizontal normal. Since it was directed downwards, its absolute direction is −30∘ relative to the positive x-axis.
The Geometric Revelation
Here is where the magic happens. Look at the face BC. The interior angle at B is 60∘. Since AB is vertical, the face BC tilts downwards at an angle of 90∘−60∘=30∘ to the horizontal.
Wait a minute! Our refracted ray is also travelling at exactly 30∘ below the horizontal. This means the refracted ray is perfectly parallel to the face BC! It will glide alongside BC without ever touching it, heading straight for the next face, CD.
Incidence and TIR at Face CD
The ray strikes face CD. To find the angle of incidence, we need the angle between the ray and the face CD. Since the ray is parallel to BC, the angle it makes with CD is simply the supplement of the interior angle at C (135∘).
Angle with surface=180∘−135∘=45∘
The angle of incidence i2 (angle with the normal) is therefore 90∘−45∘=45∘.
Will the ray escape? Let's check the critical angle θc for the prism:
sinθc=μ1=31≈0.577
Since sin45∘=21≈0.707, we clearly see that sini2>sinθc. The ray is trapped! It undergoes Total Internal Reflection (TIR) at face CD. This confirms that Option (a) is correct.
Reflection towards Face AD
Upon reflection, the ray obeys the law of reflection, turning by an angle of 180∘−2(45∘)=90∘. It was travelling at −30∘ and now takes a sharp 90∘ clockwise turn, plunging downwards and leftwards at an absolute angle of −120∘.
It heads straight for the bottom horizontal face AD. Since AD is horizontal, its normal is perfectly vertical (−90∘). The angle of incidence i3 at this face is the difference between the ray's direction and the normal:
i3=∣−120∘−(−90∘)∣=30∘
Emergence from Face AD
Since i3=30∘ is less than the critical angle, the ray will successfully refract out of the prism through face AD. This confirms that Option (b) is correct.
Let's apply Snell's Law one last time to find the angle of emergence e:
3⋅sin30∘=1⋅sine
3⋅21=sine⟹e=60∘
The ray bends away from the normal. Inside, it was 30∘ away from the vertical normal. Outside, it bends to 60∘ away from the vertical normal. Its final absolute direction becomes −90∘−60∘=−150∘.
The Final Angle
Finally, we evaluate the total deviation. The initial incident ray was directed at −60∘ to the horizontal. The final emergent ray is directed at −150∘ to the horizontal.
The angle between them is simply the absolute difference:
Angle=∣−150∘−(−60∘)∣=90∘
This confirms that Option (c) is correct, and Option (d) is incorrect. The ray has been elegantly routed through the prism, emerging exactly perpendicular to its original path!
