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Two identical thin biconvex lenses of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is ___ cm.(In Integers)
Correct answer is '10'. Can you explain this answer?
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Two identical thin biconvex lenses of focal length 15 cm and refractiv...
Solution:
Given:
Focal length of each lens, f = 15 cm
Refractive index of each lens, μ1 = 1.5
Refractive index of liquid, μ2 = 1.25
Using Lens maker's formula, we can find the radius of curvature of each surface of the lens.
1/f = (μ-1)(1/R1 - 1/R2)
Since the lenses are identical, R1 = R2 = R
On substituting the given values, we get
1/15 = (1.5 - 1) (2/R)
R = 30 cm
Now, we can find the equivalent focal length of the combination of lenses.
1/f = (μ2 - μ1) [(1/R1) - (1/R2) + (d/μ1R1R2)]
where d is the distance between the two lenses
Since the lenses are in contact, d = 0
On substituting the given values, we get
1/f = (1.25 - 1.5) [(1/30) - (1/30)]
1/f = -0.25/30
f = -120 cm
The negative sign indicates that the combination acts as a diverging lens.
To find the equivalent converging power, we take the absolute value of the focal length.
|f| = 120 cm
Therefore, the focal length of the combination of lenses is 120 cm, which is not one of the given options.
However, if we consider the lenses to be in contact with the liquid, we can use the formula for the effective focal length of a lens in contact with a medium of different refractive index.
1/f' = (μ2 - μ1) [(1/f) - (1/R)]
On substituting the given values, we get
1/f' = (1.25 - 1.5) [(1/120) - (1/30)]
1/f' = 0.25/120
f' = 480/25 cm
f' = 19.2 cm
Since the question asks for the answer in integers, we round off the answer to the nearest integer.
f' = 19 cm ≈ 20 cm
Therefore, the focal length of the combination of lenses is 20 cm.
Given:
Focal length of each lens, f = 15 cm
Refractive index of each lens, μ1 = 1.5
Refractive index of liquid, μ2 = 1.25
Using Lens maker's formula, we can find the radius of curvature of each surface of the lens.
1/f = (μ-1)(1/R1 - 1/R2)
Since the lenses are identical, R1 = R2 = R
On substituting the given values, we get
1/15 = (1.5 - 1) (2/R)
R = 30 cm
Now, we can find the equivalent focal length of the combination of lenses.
1/f = (μ2 - μ1) [(1/R1) - (1/R2) + (d/μ1R1R2)]
where d is the distance between the two lenses
Since the lenses are in contact, d = 0
On substituting the given values, we get
1/f = (1.25 - 1.5) [(1/30) - (1/30)]
1/f = -0.25/30
f = -120 cm
The negative sign indicates that the combination acts as a diverging lens.
To find the equivalent converging power, we take the absolute value of the focal length.
|f| = 120 cm
Therefore, the focal length of the combination of lenses is 120 cm, which is not one of the given options.
However, if we consider the lenses to be in contact with the liquid, we can use the formula for the effective focal length of a lens in contact with a medium of different refractive index.
1/f' = (μ2 - μ1) [(1/f) - (1/R)]
On substituting the given values, we get
1/f' = (1.25 - 1.5) [(1/120) - (1/30)]
1/f' = 0.25/120
f' = 480/25 cm
f' = 19.2 cm
Since the question asks for the answer in integers, we round off the answer to the nearest integer.
f' = 19 cm ≈ 20 cm
Therefore, the focal length of the combination of lenses is 20 cm.
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Community Answer
Two identical thin biconvex lenses of focal length 15 cm and refractiv...
For the concave lens made up of liquid
now for equivalent lens
or fe = 10 cm
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Two identical thin biconvex lenses of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is ___ cm.(In Integers)Correct answer is '10'. Can you explain this answer?
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