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Hint: By using the formula of refraction at the spherical refracting surface from rarer to the denser medium we can find the distance from the center where the child's nose appears.
Complete step by step answer:
Refraction at the spherical refracting surface from rarer to denser medium can be given by,
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$
$ \Rightarrow \dfrac{{\dfrac{4}{3}}}{v} - \dfrac{1}{{ - R}} = \dfrac{{\dfrac{4}{3} - 1}}{R}$
$ \Rightarrow \dfrac{4}{{3v}} + \dfrac{1}{R} = \dfrac{1}{{3R}}$
$ \Rightarrow \dfrac{4}{{3v}} = - \dfrac{2}{{3R}}$
$ \Rightarrow v = - \dfrac{{4R}}{2} = - 2R$
The value is, $v = 2R$
Hence the right answer is option \[(B) \Rightarrow 2R\].
Additional information:
Laws of refraction:
The incident ray, the refracted ray, and the normal at the point of incidence all lie in the same plane
$\dfrac{{\sin i}}{{\sin r}} = \mu $ [refractive index of the second medium with respect to the first medium]
Also \[\mu = \dfrac{c}{v}\]
$\mu = \dfrac{{{\text{Real depth}}}}{{{\text{Apparent depth}}}}$
Total internal reflection: When light travels from denser to rarer medium then the light incident at an angle greater than the critical angle, it is reflected back in the denser medium.
Note: The change in direction or bending of a light wave passing from one transparent medium to another medium caused by the change in wave’s speed is the Refraction.
A ray traveling along a path of the reflected ray is reflected along the path of the incident ray. In the same way, a refracted ray reversed to travel back along its path will get refracted along the path of the incident ray. So, the incident rays and refracted rays are mutually reversible.
Complete step by step answer:
Refraction at the spherical refracting surface from rarer to denser medium can be given by,
$\dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R}$
$ \Rightarrow \dfrac{{\dfrac{4}{3}}}{v} - \dfrac{1}{{ - R}} = \dfrac{{\dfrac{4}{3} - 1}}{R}$
$ \Rightarrow \dfrac{4}{{3v}} + \dfrac{1}{R} = \dfrac{1}{{3R}}$
$ \Rightarrow \dfrac{4}{{3v}} = - \dfrac{2}{{3R}}$
$ \Rightarrow v = - \dfrac{{4R}}{2} = - 2R$
The value is, $v = 2R$
Hence the right answer is option \[(B) \Rightarrow 2R\].
Additional information:
Laws of refraction:
The incident ray, the refracted ray, and the normal at the point of incidence all lie in the same plane
$\dfrac{{\sin i}}{{\sin r}} = \mu $ [refractive index of the second medium with respect to the first medium]
Also \[\mu = \dfrac{c}{v}\]
$\mu = \dfrac{{{\text{Real depth}}}}{{{\text{Apparent depth}}}}$
Total internal reflection: When light travels from denser to rarer medium then the light incident at an angle greater than the critical angle, it is reflected back in the denser medium.
Note: The change in direction or bending of a light wave passing from one transparent medium to another medium caused by the change in wave’s speed is the Refraction.
A ray traveling along a path of the reflected ray is reflected along the path of the incident ray. In the same way, a refracted ray reversed to travel back along its path will get refracted along the path of the incident ray. So, the incident rays and refracted rays are mutually reversible.
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