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Question: In the figure ABC is the cross section of a right angled prism and BCDE is the cross section of a gl...

In the figure ABC is the cross section of a right angled prism and BCDE is the cross section of a glass slab. The value of θ so that light incident normally on the face AB does not cross the face BC is (given sin1(35)=37sin-1{\rm{ }}\left( {\dfrac{3}{5}} \right){\rm{ }} = {\rm{ }}37^\circ )

A θ37\theta \le 37^\circ
B θ>37\theta > 37^\circ
C θ53\theta \le 53^\circ
D θ<53\theta < 53^\circ

Explanation

Solution

In this question, we first consider the diagram when the ray enters the prism from the face AB. With the use of Snell’s law, the angle of incidence can be calculated. Then using angle sum property, theta can be calculated.

Complete step by step answer: Given: The refractive index of the prim is μ1=32{\mu _1} = \dfrac{3}{2}, and The refractive index of the glass slab is μ2=65{\mu _2} = \dfrac{6}{5}.

We see from the diagram that there is a prism and glass slab. The beam of light is incident on the prism normally on the face AB, in such a way that the light does not cross the BC.
To find the incident angle, we have to consider that the beam of light is entered in the prism normally with the angle θ\theta
Consider the diagram when the beam entered the glass slab.

Now, to find the angle θ\theta we can use the Snell’s law,
μ1sini=μ2sinr{\mu _1}\sin i = {\mu _2}\sin r
Here, sini\sin i is the angle of incidence and sinr{\mathop{\rm sinr}\nolimits} is the angle of reflection.
Substitute the values in the above equation we get,
32sini=65sin90 sini=45 i=53 \Rightarrow \dfrac{3}{2}\sin i = \dfrac{6}{5}\sin 90^\circ \\\ \Rightarrow \sin i = \dfrac{4}{5}\\\ \Rightarrow i = 53^\circ
Therefore, the incident angle is 5353^\circ .
Now, to calculate the incident angle, we use the property of the triangle.
Therefore,
θ+90+i=180 θ+90+53=180 θ=180143 θ=37 \Rightarrow \theta + 90^\circ + i = 180^\circ \\\ \Rightarrow \theta + 90^\circ + 53^\circ = 180^\circ \\\ \Rightarrow \theta = 180^\circ - 143^\circ \\\ \Rightarrow \theta = 37^\circ
Thus, the value of theta is 3737^\circ .
The incident angle is the critical angle, so θ\theta must be less than 3737^\circ so that light does not cross the face BC.
θ37\theta \le 37^\circ

Thus, the correct option is (A).

Note: In this question, students must have the knowledge of the term reflection, refraction and critical angle. The critical angle is the angle of incidence that gives an angle of refraction of 90 degrees.