A ray of light is incident at the glass-water interface at an angle $i$ . It emerges finally parallel to the surface of water. Then the value of $μ_{g}$ would be [2003]

Question

Smart Achievers · Accepted Answer

(2)

From Snell's law, $μ_{g}$

$∴$ $μ_{g} = \frac{1}{\sin i}$

Solution

Q.
A ray of light is incident at the glass-water interface at an angle $i$ . It emerges finally parallel to the surface of water. Then the value of $μ_{g}$ would be [2003]

1 $(\frac{4}{3}) \sin i$

2 $\frac{1}{\sin i}$

3 $\frac{4}{3}$

4 $1$

Ans.
(2)

From Snell's law, $μ_{g} \sin i = μ_{air} \sin 90 °$

$∴$ $μ_{g} = \frac{1}{\sin i}$

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Solution

Q. A ray of light is incident at the glass-water interface at an angle i. It emerges finally parallel to the surface of water. Then the value of μg would be [2003]

1 (43)sini

2 1sini

3 43

4 1

Ans. (2) From Snell's law, μgsini=μairsin90° ∴ μg=1sini