Question

Radii of curvature of a biconvex lens are equal the refrective index of the material of the lens is 1.5 if the focal length of the lens is 30 cm calculate the radii of curvature of the lens

Answer 1

Explanation:

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Answer 2

Answer:

Radii of Curvatures (R) of the Biconvex lens are Equal, so R1 = R2 = R

∴ According to the Lens Formula,

= 1/f = (n - 1) [1/$R^1$ + 1/$R^2$]

= 1/f = (n - 1) [1/R + 1/R]

= 1/f = (n - 1) (2/R)

= 1/f = 2(n-1)/R

⇒ Focal Length of the Lens f = R/2(n - 1)

∵ We know that,

= 1/f = 1/u + 1/u

So now, Replace 1/f in the Lens formula

= 1/f = (n-1) [1/$R^1$ + 1/$R^2$]

= 1/u + 1/v = (n - 1) [1/R + 1/R]

= 1/u + 1/v = (n - 1) [2/R] = 2n - 2/R

= 1/u + 1/v = 2n - 2/R

Object is placed at the Centre of Curvature.

So, Object Distance (u) = R

Let: The Image formaed = V

From the above Equation,

= 1/v = 2n - 2/R - 1/u = 2(n - 1)/R - 1/R

1/v = 2n - 2 - 1/R - 2n - 3/R

∴ Image Distance = v = R/2n - 3

So, the nature of the Image is Inverted and Object Distance (v) < Image Distance (u)