
The wavelength of light in two liquids ′x′ and ′y′ is 3500A° and 7000A°. Then the critical angle of x relative to y will be :
A . 60°
B . 45°
C . 30°
D . 15°
Answer
162.6k+ views
Hint:In this question we will use the relationship between frequency and refractive index of a particular medium. We will also use the relationship between critical angle and the refractive index of a given medium.
Formula used:
Refractive index will be:
${{\mu }}=\dfrac{c}{{{v}}}$
Complete answer:
Wavelength of light in liquid x = 35000A° and in liquid y=7000A°.
We are aware that a liquid's refractive index(μ) is calculated by dividing the velocity of the light(c) by the velocity of the wave(v) in that liquid.
Refractive index of liquid x will be:
${{\mu }_{x}}=\dfrac{c}{{{v}_{x}}}$ - (1)
Refractive index of liquid y will be:
${{\mu }_{y}}=\dfrac{c}{{{v}_{y}}}$ - (2)
Dividing equation (1) by equation (2)
$\dfrac{{{\mu }_{x}}}{{{\mu }_{y}}}=\dfrac{{{v}_{y}}}{{{v}_{x}}}$ - (3)
For a wave travelling in a straight line, the wavelength can be calculated by the division of velocity(v) by frequency of wave(f) of the wave.
$\lambda =\dfrac{v}{f}$
$\Rightarrow v=f\lambda $
The velocity of light in both the liquids:
${{v}_{x}}=f{{\lambda }_{x}}$
${{v}_{y}}=f{{\lambda }_{y}}$
Substituting the equation of velocity in equation (3)
$\dfrac{{{n}_{x}}}{{{n}_{y}}}=\dfrac{f{{\lambda }_{y}}}{f{{\lambda }_{x}}}$
$\Rightarrow \dfrac{{{n}_{x}}}{{{n}_{y}}}=\dfrac{{{\lambda }_{y}}}{{{\lambda }_{x}}}$ - (4)
We know that critical angle($\theta $) of medium x relative to y is:
$\theta ={{\sin }^{-1}}\dfrac{1}{\dfrac{{{n}_{x}}}{{{n}_{y}}}}$
$\Rightarrow \theta ={{\sin }^{-1}}\dfrac{1}{\dfrac{{{\lambda }_{y}}}{{{\lambda }_{x}}}}$
Substituting the values in the above expression:
$\theta ={{\sin }^{-1}}\dfrac{1}{\dfrac{7000}{3500}}$
$\Rightarrow \theta ={{\sin }^{-1}}\dfrac{1}{2}$
$\Rightarrow \theta ={{30}^{\circ }}$
Therefore, the critical angle of liquid x with respect to y is 30°.
The correct answer is C.
Note:The angle of incidence, for which the angle of refraction is 90°, is the critical angle. Light changes direction and bends toward the normal if it enters a denser medium from a rarer medium. Beyond the critical angle, all the waves reflect back into the glass and we say that the rays are totally internally reflected.
Formula used:
Refractive index will be:
${{\mu }}=\dfrac{c}{{{v}}}$
Complete answer:
Wavelength of light in liquid x = 35000A° and in liquid y=7000A°.
We are aware that a liquid's refractive index(μ) is calculated by dividing the velocity of the light(c) by the velocity of the wave(v) in that liquid.
Refractive index of liquid x will be:
${{\mu }_{x}}=\dfrac{c}{{{v}_{x}}}$ - (1)
Refractive index of liquid y will be:
${{\mu }_{y}}=\dfrac{c}{{{v}_{y}}}$ - (2)
Dividing equation (1) by equation (2)
$\dfrac{{{\mu }_{x}}}{{{\mu }_{y}}}=\dfrac{{{v}_{y}}}{{{v}_{x}}}$ - (3)
For a wave travelling in a straight line, the wavelength can be calculated by the division of velocity(v) by frequency of wave(f) of the wave.
$\lambda =\dfrac{v}{f}$
$\Rightarrow v=f\lambda $
The velocity of light in both the liquids:
${{v}_{x}}=f{{\lambda }_{x}}$
${{v}_{y}}=f{{\lambda }_{y}}$
Substituting the equation of velocity in equation (3)
$\dfrac{{{n}_{x}}}{{{n}_{y}}}=\dfrac{f{{\lambda }_{y}}}{f{{\lambda }_{x}}}$
$\Rightarrow \dfrac{{{n}_{x}}}{{{n}_{y}}}=\dfrac{{{\lambda }_{y}}}{{{\lambda }_{x}}}$ - (4)
We know that critical angle($\theta $) of medium x relative to y is:
$\theta ={{\sin }^{-1}}\dfrac{1}{\dfrac{{{n}_{x}}}{{{n}_{y}}}}$
$\Rightarrow \theta ={{\sin }^{-1}}\dfrac{1}{\dfrac{{{\lambda }_{y}}}{{{\lambda }_{x}}}}$
Substituting the values in the above expression:
$\theta ={{\sin }^{-1}}\dfrac{1}{\dfrac{7000}{3500}}$
$\Rightarrow \theta ={{\sin }^{-1}}\dfrac{1}{2}$
$\Rightarrow \theta ={{30}^{\circ }}$
Therefore, the critical angle of liquid x with respect to y is 30°.
The correct answer is C.
Note:The angle of incidence, for which the angle of refraction is 90°, is the critical angle. Light changes direction and bends toward the normal if it enters a denser medium from a rarer medium. Beyond the critical angle, all the waves reflect back into the glass and we say that the rays are totally internally reflected.
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