An equiconvex lens 350 cm thick has radii of curvature of...

An equiconvex lens 3.50 cm thick has radii of curvature of 5.20 cm and is made of a glass with refractive index n = 1.680. Calculate (a) the focal length of the lens and (b) the power of the lens. Obtain the positions of the principal planes and the distances of the vertices to the respective foci.

Transcript

00:01 Here in this problem we have been given a thick lens and using lens maker's formula we have to find focal length and hence the power of this thick lens this is the thickness of the lens let it be deep and then we will find its focal planes f each on both the sides so in the first part of this problem focal length of the thick lens is given by lens makers formula as one by f inverse of its focal length is equal to n minus 1 where n is the index of refraction of this material of the lens bracket 1 by r1 radius of curvature of the left side surface minus 1 by r2 radius of curvature of the right side surface r2 and plus n minus 1 1, again index of refraction into d, thickness of the lens and divided by n into r1 into r2.

01:52 So finally plugging in all the known values here, index of refraction has been given as 1 .680 minus 1.

02:03 As per sign convention, this 1 by r1 radius of curvature, both of the sides are having same radii because it is an equi convex lens so this is 5 .20 centimeter for r1 but for r2 as per sign convention it will become negative so this is minus minus 5 .20 plus for n again 1 .680 minus 1 minus 1 into thickness of the lens which is 3 .50 centimeter divided by n again means 1 .680 for r1 5 .20 for r2 minus 5 .20 centimeter now here it is 0 .680 outside the bracket in the bracket here it will be positive so it will become plus so these are 2 divided by 5 .20 and here it will become minus negative because of this negative sign so minus this is 0 .680 multiplied by 3 .50 and it will come out to be 1 .42 divided by the product of these three terms and here it is 5 .20 square actually this 1 .4 2 is 0 .680 multiplied by 3 .50 and divided by 1 .60 0 .680 leaving behind to 5 .20 product of 5 .20 which will become square of 5 .20.

03:47 So now taking lcm and solving it further we get 1 by f inverse of the focal length to be equal to 0 .2 3 or focal length of this thick convex lens is 4 .43 cm which is answer for the first part of this problem.

04:13 Now in the second part of the problem we have to find power of this converging lens.

04:23 That power is given as inverse of focal length but it should be taken in meters...

Question

An equiconvex lens 3.50 cm thick has radii of curvature of 5.20 cm and is made of a glass with refractive index n = 1.680. Calculate (a) the focal length of the lens and (b) the power of the lens. Obtain the positions of the principal planes and the distances of the vertices to the respective foci.

Please give Ace some feedback