Question

The refractive index of convex lens material is 1.46 and the refractive index of Benzene and water is 1.5 and 1.0 respectively. How does the lens behaves when it is kept in Benzene and water ? Given and write​

Answer 1

Explanation:

We know that the formula for focal length of a lens is given by,

$\dfrac{1}{f} = ( \mu - 1)( \dfrac{1}{r1} - \dfrac{1}{r2} )$

Where,

• f is focal length
• μ is refractive index of the lens
• r1 and r2 are the radii of curvatures

Now, we also know that convex lens is converging lens because focal length of convex lens is positive.

So, power of lens will also be positive.

Now, when the lens is placed in Benzene of refractive index 1.50

Then we will get,

=> μ = refractive index of lens/refractive index of benzene

=> μ = 1.46/1.150

Clearly, we have, 1.46/1.50 < 1

=> μ < 1

=> (μ - 1 )< 0

Therefore, from the above focal length formula,

It is now clear that 1/f < 0 when (μ - 1 )< 0. Thus, focal length will be negative.

But, we know that 1/f is the power.

Therefore, power will also be negative .

Hence, lens will be diverging in benzene liquid.

Now, if the lens is immersed in water,

Then we will have,

=> μ = 1.50/1

But, 1.5 > 1

Therefore, we will have,

=> μ > 1

=> μ - 1 > 0

Now, from the above formula,

It's clear that,

Focal length will be positive.

Thus, the power of lens will also be positive.

Hence, nature of lens will remain unchanged, i.e., converging lens.

Answer 2

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• μ is relative refractive index of lens.

• Means refractive index of lens with respect to medium.

• You also know , convex lens is converging lens because focal length of convex lens is positive.

• So, power of lens will be positive.

• Now, when lens is placed in Benzene of refractive index 1.50

• Then, Relative refractive index of lens,

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μ = $\frac{refractive\: index \:of \:lens}{refractive \:index\: of \:benzene }$

μ = $\huge \frac{1.46}{1.150}$ < 1

(μ - 1 ) < 0 from equation

$\frac{1}{F}$ < 0 when (μ - 1 ) < 0

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• Hence , focal length will be negative. So, power will be negative .

• Hence, Lens will be diverging in benzene liquid.

• Now , if lens is immersed in water then , relative refractive index of lens = 1.50/1 > 1

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• Focal length positive.
• So, power of lens will be positive.
• Hence, nature of lens unchange.
• e.g. → Converging.

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