The focal length of an equiconvex lens is equal to the radius of curvature of either face .What is the refractive index of the lens material?
Answers
Answer:Its focal length will be 3/2
Explanation:
As it is an equiconvex lens radius of curvature of both sides will be equal to R.
Also f=R (given)
Using lens makers formula;
1/f= (m-1){1/R - (-1/R)}
Since f = R,
1/R= (m-1)×2/R
1/2= m-1
m = 3/2.
The refractive index of the lens material is 3/2. This means that the material has a higher index of refraction than air, which has a refractive index of approximately 1.
An equiconvex lens is a type of lens where both faces have the same radius of curvature (R). If the focal length (f) of the lens is equal to the radius of curvature (R), then the lens can be described using the lens maker's formula.
The lens maker's formula states that:
1/f = (m - 1) × (1/R - (-1/R))
where m is the refractive index of the lens material.
If f = R, then we can simplify the formula to:
1/R = (m - 1) × 2/R
Solving for m, we find that:
m = (1/2) + 1 = 3/2
In conclusion, when the focal length of an equiconvex lens is equal to the radius of curvature of either face, the refractive index of the lens material can be calculated using the lens maker's formula. The refractive index is a measure of how much a material bends light, and is an important characteristic of lens materials.
For more such questions on the refractive index: https://brainly.in/question/23618802
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