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Glass has refractive index mu with respe...

Glass has refractive index `mu` with respect to air and the critical angle for a ray of light going from glass to air is `theta`. If a ray of light is incident from air on the glass with angle of incidence `theta`, the corresponding angle of refraction is

A

`sin^(-1)((1)/(sqrt(mu)))`

B

`90^(@)`

C

`sin^(-1)((1)/(mu^(2)))`

D

`sin^(-1)((1)/(mu))`

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The correct Answer is:
To solve the problem, we will use Snell's law, which relates the angles of incidence and refraction to the refractive indices of the two media involved. ### Step-by-Step Solution: 1. **Understanding the Critical Angle**: The critical angle \( \theta \) is defined as the angle of incidence in the denser medium (glass) that results in the angle of refraction being \( 90^\circ \) in the less dense medium (air). According to Snell's law: \[ \mu \sin \theta = 1 \cdot \sin 90^\circ \] Since \( \sin 90^\circ = 1 \), we can simplify this to: \[ \mu \sin \theta = 1 \] This gives us: \[ \mu = \frac{1}{\sin \theta} \] 2. **Setting Up the Problem**: Now, we need to find the angle of refraction \( r \) when a ray of light is incident from air into glass at the angle \( \theta \). According to Snell's law for this scenario: \[ n_1 \sin \theta = n_2 \sin r \] Here, \( n_1 = 1 \) (refractive index of air), \( n_2 = \mu \) (refractive index of glass), and \( \theta \) is the angle of incidence. 3. **Applying Snell's Law**: Substituting the values into Snell's law: \[ 1 \cdot \sin \theta = \mu \sin r \] Rearranging gives: \[ \sin r = \frac{\sin \theta}{\mu} \] 4. **Substituting for \( \sin \theta \)**: From the earlier step, we know that \( \mu = \frac{1}{\sin \theta} \). Therefore, we can express \( \sin \theta \) as: \[ \sin \theta = \frac{1}{\mu} \] Substituting this back into our equation for \( \sin r \): \[ \sin r = \frac{\frac{1}{\mu}}{\mu} = \frac{1}{\mu^2} \] 5. **Finding the Angle of Refraction**: To find the angle \( r \), we take the inverse sine: \[ r = \sin^{-1} \left( \frac{1}{\mu^2} \right) \] ### Final Answer: The corresponding angle of refraction \( r \) is: \[ r = \sin^{-1} \left( \frac{1}{\mu^2} \right) \]

To solve the problem, we will use Snell's law, which relates the angles of incidence and refraction to the refractive indices of the two media involved. ### Step-by-Step Solution: 1. **Understanding the Critical Angle**: The critical angle \( \theta \) is defined as the angle of incidence in the denser medium (glass) that results in the angle of refraction being \( 90^\circ \) in the less dense medium (air). According to Snell's law: \[ \mu \sin \theta = 1 \cdot \sin 90^\circ ...
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A2Z-GEOMETRICAL OPTICS-Section D - Chapter End Test
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  2. Wavelength of light used in an optical instrument are lambda(1)=400 Å ...

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  3. A plano convex lens of refractive index 1.5 and radius of curvature 30...

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  4. A light ray is incident perpendicularly to one face of a 90^circ prism...

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  5. A thin glass (refractive index 1.5) lens has optical power of -5D in a...

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  6. Which of the following graphs is the magnification of a real image aga...

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  7. A thin prism P(1) with angle 4degree and made from glass of refractive...

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  8. A converging lens is used to form an image on a screen. When the upper...

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  9. A diminished image of an object is to be obtained on a screen 1.0 m fr...

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  10. An object 15cm high is placed 10cm from the optical center of a thin l...

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  11. A lens forms a virtual, diminished image of an object placed at 2 m fr...

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  12. When the distance between the object and the screen is more than 4 f....

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  13. a convex lens of power +6 diopter is placed in contact with a concave ...

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  14. A concave lens of focal length 20 cm product an image half in size of ...

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  15. A convex lens of focal length 1.0m and a concave lens of focal length ...

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  16. If in a planoconvex lens, the radius of curvature of the convex surfac...

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  17. A convex lens A of focal length 20cm and a concave lens G of focal le...

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  18. The radii of curvature of the two surfaces of a lens are 20cm and 30 c...

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  19. A lens forms a virtual image 4 cm away from it when an object is place...

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  20. A concave lens of focal length (1)/(3)m forms a real, inverted image t...

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  21. An object is placed at a distance of f//2 from a convex lens. The imag...

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