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# A lean boy Ramesh placed a glass slab of refractive index $\dfrac{3}{2}$on his photograph. He felt it more interesting as his face appeared bigger than the original size when observed from the top of the glass slab. He also observed that the photo appears at a height of 1.5cm from the bottom. He then placed another glass cube made of different material and having the same thickness over the first glass cube. Now, he himself started to laugh by seeing his photograph as it appeared still bigger than the previous. When observed from the top the photograph appeared to be at a height 4cm from the bottom. Determine the refractive index of the second glass cube.A) 1.5B) 2.25C) 3.75D) 1.75

Last updated date: 14th Sep 2024
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Hint: We have all done this in our childhood taking a magnifying glass/lens and showing our face from it, by doing this we observed that our face looks bigger than usual. This is the play of light passing through the glass/lens and getting refracted. The light is vertically shifted due to the variation in the refractive index (medium). The light goes from a rarer medium (air) to a denser medium (glass).

Formula Used:
The relation between thickness of the material, refractive index of the glass and the height of the image formed is given as:
$\delta = t\left( {\mu - 1} \right)$;
Where
$\delta$= Height of the image formed;
$\mu$= refractive index of the glass;
t = Thickness of the material

The shift (vertically) due to variation in refractive index is given by:
$\delta = t\left( {\mu - 1} \right)$;
So, we have 2 objects, write the above equation in terms of the two objects:
${\delta _1} = t\left( {{\mu _1} - 1} \right)$; …(1)
${\delta _2} = t\left( {{\mu _2} - 1} \right)$; …(2)
Find the refractive index of the second glass object${\mu _2}$:
Divide the 1st equation with the 2nd equation and solve:
$\dfrac{{{\delta _1}}}{{{\delta _2}}} = \dfrac{{t\left( {{\mu _1} - 1} \right)}}{{t\left( {{\mu _2} - 1} \right)}}$ ;
$\Rightarrow \dfrac{{{\delta _1}}}{{{\delta _2}}} = \dfrac{{\left( {{\mu _1} - 1} \right)}}{{\left( {{\mu _2} - 1} \right)}}$
Put in the given values in the above equation:
$\dfrac{{1.5}}{4} = \dfrac{{\left( {\dfrac{3}{2} - 1} \right)}}{{\left( {{\mu _2} - 1} \right)}}$;
$\Rightarrow \dfrac{{1.5}}{4} \times \left( {{\mu _2} - 1} \right) = \left( {\dfrac{3}{2} - 1} \right)$;
Simplify the above equation:
$\Rightarrow \dfrac{{1.5}}{4} \times \left( {{\mu _2} - 1} \right) = \left( {\dfrac{{3 - 2}}{2}} \right)$;
$\Rightarrow \dfrac{{1.5}}{2} \times \left( {{\mu _2} - 1} \right) = 1$;
Solve,
$\Rightarrow \left( {{\mu _2} - 1} \right) = \dfrac{2}{{1.5}}$;
$\Rightarrow {\mu _2} = \dfrac{2}{{1.5}} + 1$;
The refractive index of the second glass object is:
$\Rightarrow {\mu _2} \simeq 2.25$;
$\therefore$ The correct option is B

Note: Here we have to establish a relation between the height, thickness and the refractive index of the medium, for two glass objects of the same thickness ‘t’. After that divide the first relation with the second relation and find out the refractive index for the second medium.