
A lens is made of flint glass (having refractive index= 1.5). When this lens is immersed in a liquid of refractive index 1.25, the focal length;
A. Increases to a factor of 1.25
B. Increases to a factor of 2.5
C. Increases to a factor of 1.2
D. Decreases to the factor of 1.2
Answer
213.9k+ views
Hint: As in the question they are talking about the relation between refractive index and focal length so we use lens maker formula i.e. $\dfrac{1}{f} = \left( {\mu - 1} \right)\left[ {\dfrac{1}{{r_1}} - \dfrac{1}{{r_2}}} \right]$. Firstly, write the lens maker formula for both cases, one is when the lens is placed in air and when the lens is immersed in water. After that divide both equations we will get the ratio between both focal lengths which is the required increase in the focal length.
Complete step-by-step answer:
As, it is given that refractive index of flint glass (µg) with respect to air = 1.5
And the refractive index of liquid (µ) =1.25
Now, the lens maker formula with respect to air$\dfrac{1}{{fg}} = \left( {\mu g - 1} \right)\left[ {\dfrac{1}{{r_1}} - \dfrac{1}{{r_2}}} \right]$ ………………………………… (1)
⇒$\dfrac{1}{{fg}} = \left( {1.5 - 1} \right)\left[ {\dfrac{1}{{r_1}} - \dfrac{1}{{r_2}}} \right]$……………………….. (2)
Now when we immersed in water then formula is $\dfrac{1}{{fl}} = \left( {\mu l - 1} \right)\left[ {\dfrac{1}{{r_1}} - \dfrac{1}{{r_2}}} \right]$…………………… (3)
Now, the refractive of glass with respect to liquid is $\mu l = \dfrac{{\mu g}}{\mu } = \dfrac{{1.5}}{{1.25}} = \dfrac{6}{5}$
Put this value in above equation (2), we get $\dfrac{1}{{fl}} = \left( {\dfrac{6}{5} - 1} \right)\left[ {\dfrac{1}{{r_1}} - \dfrac{1}{{r_2}}} \right] = \dfrac{1}{5}\left( {\dfrac{1}{{r_1}} - \dfrac{1}{{r_2}}} \right)$………………. (4)
Now, divide eq (2) by eq (3), we get
⇒$\dfrac{{fl}}{{fg}} = 0.5 \times 5 = 2.5$
Hence, focal length of 2.5. Therefore, B option is correct.
Note: The refractive index is the ratio of speed of light in the medium 1 to the speed of light in medium 2. Remember to take the refractive index of glass with respect to liquid in the equation 4 because in this case also light from the air medium into the water medium, so care must be taken otherwise the answer would be incorrect.
Complete step-by-step answer:
As, it is given that refractive index of flint glass (µg) with respect to air = 1.5
And the refractive index of liquid (µ) =1.25
Now, the lens maker formula with respect to air$\dfrac{1}{{fg}} = \left( {\mu g - 1} \right)\left[ {\dfrac{1}{{r_1}} - \dfrac{1}{{r_2}}} \right]$ ………………………………… (1)
⇒$\dfrac{1}{{fg}} = \left( {1.5 - 1} \right)\left[ {\dfrac{1}{{r_1}} - \dfrac{1}{{r_2}}} \right]$……………………….. (2)
Now when we immersed in water then formula is $\dfrac{1}{{fl}} = \left( {\mu l - 1} \right)\left[ {\dfrac{1}{{r_1}} - \dfrac{1}{{r_2}}} \right]$…………………… (3)
Now, the refractive of glass with respect to liquid is $\mu l = \dfrac{{\mu g}}{\mu } = \dfrac{{1.5}}{{1.25}} = \dfrac{6}{5}$
Put this value in above equation (2), we get $\dfrac{1}{{fl}} = \left( {\dfrac{6}{5} - 1} \right)\left[ {\dfrac{1}{{r_1}} - \dfrac{1}{{r_2}}} \right] = \dfrac{1}{5}\left( {\dfrac{1}{{r_1}} - \dfrac{1}{{r_2}}} \right)$………………. (4)
Now, divide eq (2) by eq (3), we get
⇒$\dfrac{{fl}}{{fg}} = 0.5 \times 5 = 2.5$
Hence, focal length of 2.5. Therefore, B option is correct.
Note: The refractive index is the ratio of speed of light in the medium 1 to the speed of light in medium 2. Remember to take the refractive index of glass with respect to liquid in the equation 4 because in this case also light from the air medium into the water medium, so care must be taken otherwise the answer would be incorrect.
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