
A glass lens is dipped inside a liquid medium which has a refractive index same as that of a glass lens. So what will happen to its focal length?
Answer
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Hint: The refractive index of a medium is described as how the light ray travels via that considered medium. It is a dimensionless measurement. It defines how much a light ray may be deviated when it enters from one medium to the other medium. Snell’s regulation clarifies that the relation among the angle of incidence and angle of refraction. So, there are formulations for calculating the refractive index of a medium.
Therefore by using snell’s law,
\[n = \dfrac{{\sin i}}{{\sin r}}\]
Where, n is the refractive index and ratio of incidence i, reflective r angle.
Complete step by step solution:
To find focal length as it is inversely proportional to refractive index of that medium and for we have formula for two different mediums,
\[\dfrac{1}{f} = \left( {n - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]
Here f is the focal length, n is the refractive index and ${R_1}$ and ${R_2}$ are the radius of curvature. From the question it is clear that both radii can be taken as equal.
Now, substitute all the given values in this formula, we get-
\[\dfrac{1}{f} = \left( {n - 1} \right) \times 0\]
\[\Rightarrow f = \dfrac{1}{0}\]
\[\Rightarrow f = \infty \]
Here focal length is infinite.
Therefore, the refractive index of a lens will have a decrease when it is dipped inside a denser medium than that of air, hence focal length will have an increase.
A glass lens dipped in water will behave like a plane glass plate. Focal length of the lens will emerge as infinity and the power will emerge as zero. Therefore, it will not behave like a lens.
Note: When you dip any lens or glass slab in a liquid whose refractive index n is equal to that of lens or glass slab will be similar as liquid as it was before dipping Refractive index of lens will approximately equal to 1 in that liquid and focal length will become infinity, going through the formulas for focal length as studied above.
Therefore by using snell’s law,
\[n = \dfrac{{\sin i}}{{\sin r}}\]
Where, n is the refractive index and ratio of incidence i, reflective r angle.
Complete step by step solution:
To find focal length as it is inversely proportional to refractive index of that medium and for we have formula for two different mediums,
\[\dfrac{1}{f} = \left( {n - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]
Here f is the focal length, n is the refractive index and ${R_1}$ and ${R_2}$ are the radius of curvature. From the question it is clear that both radii can be taken as equal.
Now, substitute all the given values in this formula, we get-
\[\dfrac{1}{f} = \left( {n - 1} \right) \times 0\]
\[\Rightarrow f = \dfrac{1}{0}\]
\[\Rightarrow f = \infty \]
Here focal length is infinite.
Therefore, the refractive index of a lens will have a decrease when it is dipped inside a denser medium than that of air, hence focal length will have an increase.
A glass lens dipped in water will behave like a plane glass plate. Focal length of the lens will emerge as infinity and the power will emerge as zero. Therefore, it will not behave like a lens.
Note: When you dip any lens or glass slab in a liquid whose refractive index n is equal to that of lens or glass slab will be similar as liquid as it was before dipping Refractive index of lens will approximately equal to 1 in that liquid and focal length will become infinity, going through the formulas for focal length as studied above.
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