Question

An equiconvex lens of focal length f is cut into two
identical plane convex lenses. How will the power of
each part be related to the focal length of the original
lens?
A double convex lens of + 5D is made of glass of
refractive index 1.55 with both faces of equal radii of
curvature. Find the value of its radius of curvature.
IT​

Answer 1

Answer:

Focal Length of each part is half of the original focal length. Radius of Curvature is 40 cm

Explanation:

a)  If an Equiconvex lens is cut into two equal parts, the focal length of each part is half of the original focal length.

If the focal length of the initial lens is F. The focal length of the new lens are       F/2.

b)  Power = 1/ Focal length

    Focal length = 1/ Power

   ⇒ f = 1/5 = 20cm

   Also, 2×Focal length = Radius of curvature

   ⇒  Radius of curvature = 40cm.

Hope my answer helped you :)