how to solve sum when a face of lens is silvered ? i used the formula for focal length of silvered lens but i dint get the answer ?
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We use the formula for refraction through the curved surface (convex) of the lens, when the light falls on the unsilvered face of the lens.
We assume that R = radius of curvature of either surface of convex lens. μ_g = refractive index of the material of lens. u = object distance.
1/v1 - 1/u1 = (μ_g - 1) / R --- (1)
1/v1 = (μ_g - 1) / R - 1/u1 --- (2) , using sign convention:
From the above we find the image distance.
The rays from the object get refracted and fall on the silvered surface of the lens. We should treat the silvered surface as a concave mirror. Hence, the light rays are reflected and converge at the image distance. we may get a real inverted image.
v1 from the above calculation, will act as the virtual object distance for the concave mirror. We should use the appropriate sign as per sign convention.
Then focal length of concave mirror will be equal to R/2 where R = radius of curvature.
so we use the formula: 1/v + 1/u = 1/f ------ (3)
We find v as we know u and f. and sign convention.
1/v2 - 1/v1 = 2/R
1/v2 = 2 / R + (μ_g - 1) / R - 1/u1
1 / v2 + 1 / u1 = (1 + μ_g) / R
This is equivalent to a concave mirror with a focal length : (1 + μ_g) / R
So the focal length is smaller than R / 2.
I hope it is correct, I am not sure.
We assume that R = radius of curvature of either surface of convex lens. μ_g = refractive index of the material of lens. u = object distance.
1/v1 - 1/u1 = (μ_g - 1) / R --- (1)
1/v1 = (μ_g - 1) / R - 1/u1 --- (2) , using sign convention:
From the above we find the image distance.
The rays from the object get refracted and fall on the silvered surface of the lens. We should treat the silvered surface as a concave mirror. Hence, the light rays are reflected and converge at the image distance. we may get a real inverted image.
v1 from the above calculation, will act as the virtual object distance for the concave mirror. We should use the appropriate sign as per sign convention.
Then focal length of concave mirror will be equal to R/2 where R = radius of curvature.
so we use the formula: 1/v + 1/u = 1/f ------ (3)
We find v as we know u and f. and sign convention.
1/v2 - 1/v1 = 2/R
1/v2 = 2 / R + (μ_g - 1) / R - 1/u1
1 / v2 + 1 / u1 = (1 + μ_g) / R
This is equivalent to a concave mirror with a focal length : (1 + μ_g) / R
So the focal length is smaller than R / 2.
I hope it is correct, I am not sure.
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