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Question

Consider an equilateral prism (refractive index $\sqrt{2}$). A ray of light is incident on its one surface at a certain angle $i$. If the emergent ray is found to graze along the other surface then the angle of refraction at the incident surface is close to _________.

The correct answer is
$15^\circ$

Prism Parameters and Given Condition

The problem involves an equilateral prism, which means the prism angle is $A = 60^\circ$. The refractive index of the prism is given as $n = \sqrt{2}$. A ray of light is incident on one surface, and the emergent ray grazes along the other surface. This specific condition implies that the angle of emergence is $e = 90^\circ$.

Snell's Law at the Emergent Surface

Let $r_2$ be the angle of refraction inside the prism at the second surface. According to Snell's Law at the point where the ray exits the prism:

$ n_{prism} \sin(r_2) = n_{air} \sin(e) $

Substituting the known values ($n_{prism} = \sqrt{2}$ and $e = 90^\circ$):

$ \sqrt{2} \sin(r_2) = 1 \times \sin(90^\circ) $

Since $\sin(90^\circ) = 1$:

$ \sqrt{2} \sin(r_2) = 1 $

Calculating Second Refraction Angle ($r_2$)

Solving for $\sin(r_2)$:

$ \sin(r_2) = \frac{1}{\sqrt{2}} $

Therefore, the angle $r_2$ is:

$ r_2 = 45^\circ $

Prism Angle Relationship

For any prism, the prism angle $A$ is related to the angles of refraction at the first ($r_1$) and second ($r_2$) surfaces by the formula:

$ A = r_1 + r_2 $

Calculating First Refraction Angle ($r_1$)

We need to find the angle of refraction at the incident surface, which is $r_1$. Using the prism angle formula and the calculated value of $r_2$:

$ 60^\circ = r_1 + 45^\circ $

Solving for $r_1$:

$ r_1 = 60^\circ - 45^\circ $

$ r_1 = 15^\circ $

Final Answer

The angle of refraction ($r_1$) at the incident surface is $15^\circ$.

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Similar Questions

  1. The magnitudes of power of a biconvex lens (refractive index $1.5$) and that of a plano-concave lens (refractive index = $1.7$) are same. If the curvature of plano-concave lens exactly matches with the curvature of back surface of the biconvex lens, then ratio of radius of curvature of front and back surface of the biconvex lens is ________.
  2. Given below are two statements:

    **Statement I:** A plane wave after passing through prism remains as plane wave but passing through small pin hole may become spherical wave.

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Important Questions from Optics

  1. The magnitudes of power of a biconvex lens (refractive index $1.5$) and that of a plano-concave lens (refractive index = $1.7$) are same. If the curvature of plano-concave lens exactly matches with the curvature of back surface of the biconvex lens, then ratio of radius of curvature of front and back surface of the biconvex lens is ________.
  2. Given below are two statements:

    **Statement I:** A plane wave after passing through prism remains as plane wave but passing through small pin hole may become spherical wave.

    **Statement II:** The curvature of a spherical wave emerging from a slit will increase for increasing slit width.

    In the light of the above statements, choose the correct answer from the options given below
  3. A convex lens of refractive index $1.5$ and focal length $f = 18 \text{ cm}$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times f$. The value of $\alpha$ is ________.
    (refractive index of water = $4/3$)
  4. A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification $m_1$ when an object is placed 10 cm before the convex lens. Keeping the positions of convex lens and object undisturbed a gap of 1 cm is introduced between the lenses by moving the concave lens away, which lead to a change in magnification of total lens system to $m_2$. The value of $\left| \frac{m_1}{m_2} \right|$ is _________.
  5. A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm. Refractive index of glass is 1.5. The rays converge to a point at a distance x cm from the centre of the curvature of the spherical surface. The value of x is _________ cm.
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