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Refraction At Spherical Surface MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • 17 Questions around this concept.

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An empty spherical flask of diameter 15 cm is placed in water of refractive index $4 / 3$. A parallel ray of light strikes the flask. Where does it get focused, if observed from within the flask, when measured from the centre of the surface of the spherical flask surface i.e., pole in cm

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Refraction At Spherical Surface

If an object 0 is placed in front of a curved surface as shown in the above figure, then the Refraction formula is given as

\frac{n _{2}}{v}-\frac{n _{1}}{u}= \frac{n _{2}-n_{1}}{R}

where 

 n_1=Refractive index of the medium from which light rays are coming (from the object).

n_2=Refractive index of the medium in which light rays are entering.

and n_1<n_2

and u = Distance of object, v = Distance of image, R = Radius of curvature

Note - 

  • use sign convention while solving the problem
  • Real image forms on the side of a refracting surface that is opposite to the object,
    and virtual image forms on the same side as the object.
  • Using R=\infty (i.e for plane surface)

      we get \frac{n_2}{v}=\frac{n_1}{u}\Rightarrow \frac{n_2}{n_1}=\frac{v}{u}

 

Lateral Magnification For Refracting Spherical Surface

If an object AB is placed in front of a curved surface as shown in the above figure, then the lateral Magnification formula is given as

\text { Lateral magnification, } m=\frac{\text { Image height }}{\text { Object height }}=\frac{-\left(A^{\prime} B^{\prime}\right)}{A B}

or m=-\frac{A^{\prime} B^{\prime}}{A B}=-\frac{\mu_{1}}{\mu_{2}} \times \frac{v}{u}=-\frac{v / \mu_{2}}{u / \mu_{1}}

where

 \mu _1=Refractive index of the medium from which light rays are coming (from the object).

\mu _2=Refractive index of the medium in which light rays are entering.

and \mu _1< \mu _2

and u = Distance of object, v = Distance of image, R = Radius of curvature

 

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