NEET-XII-Physics
18: Geometrical Optics
- #10Suppose you are inside the water in a swimming pool near an edge. A friends is standing on the edge. Do you find your friend taller or shorter than his usual height?Ans : When viewed from the water, the friend will seem taller than his usual height.
Let actual height be h and the apparent height be h'.
Here, the refraction is taking place from rarer to denser medium and a virtual image is formed.
Using
`` \frac{{\mu }_{1}}{-u}+\frac{{\mu }_{2}}{v}=\frac{{\mu }_{2}-{\mu }_{1}}{R}``
Where refractive index of water is μ2 and refractive index of air is μ1.
u and v are object and image distances, respectively.
R is the radius of curvature, here we will take it as ∞.
`` \frac{{\mu }_{1}}{-u}+\frac{{\mu }_{2}}{v}=\frac{{\mu }_{2}-{\mu }_{1}}{\infty }``
`` \frac{{\mu }_{1}}{u}=\frac{{\mu }_{2}}{v}``
`` v=\frac{{\mu }_{2}}{{\mu }_{1}}\times u``
As μ1 = 1
v = u × μ2
We know magnification is given by:
`` m=\frac{v}{u}``
Putting the value of v in the above equation:
`` m=\frac{u\times {\mu }_{2}}{u}``
`` m={\mu }_{2}``
`` ``
As the magnification is greater than 1, so the apparent height seems to be greater than actual height.
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