The radii of curvature of the surfaces of a thin lens are +10cm and +30 cm. The index is 1.50.
(a) Compute the position and size of the image of an object in the form of an arrow 1 cm high, perpendicular to the lens axis, 40 cm to the left of the lens.
(b) A second similar lens is placed 160 cm to the right of the first. Find the position of the final image.
(c) Same as (b) except the second lens is 40 cm to the right of the first.
(d) Same as (c) except the second lens is diverging, of focal length -40cm.
SOLUTION This solution is FREE courtesy of BrainMass!
Solution: We use the sign convention: i) All distances are measured from the pole of the lens and ii) all distances measured in the direction of the incident ray are positive and those directed opposite are negative.
We note that the radii of curvature of both the surfaces of the lens are positive. As per the sign convention: i) positive radius of curvature of the first surface implies it is convex and ii) positive radius of curvature of the second surface implies it is concave. Hence, the lens is a convexo-concave one.
First we compute the focal length of the lens using the lens maker's formula.
Lens maker's formula: 1/f = (n - 1)[1/R1 - 1/R2]
Substituting n = 1.5, R1 = +10 cm and R2 = +30 cm we get:
1/f = (1.5 - 1)[1/10 - 1/30] = 0.0333
=> f = +30 cm
a) We can compute the required values using the lens formula viz. 1/v - 1/u = 1/f
Given: Distance of the object from the lens = u = - 40 cm (-ve sign is on account of the sign convention)
Substituting values in the lens formula we get: 1/v - (-1/40) = 1/30
1/v = 1/30 - 1/40 = 1/120 => v = +120 cm
Hence, the image if formed 120 cm to the right of the lens. The image is real (as v is positive) and inverted (as it is real).
Magnification m = v/u = - 120/40 = - 3
Hence, size of the image = 3 cm
b) The image due to the first becomes the object for the second. As the second lens is placed 160 cm to the right of the first and the image due to the first lens is located 120 cm to its right, the distance of the image due to the first lens (which is the object for the second lens) from the second lens is 40 cm to its left. Hence, for the second lens u2 = - 40 cm.
As the second lens is identical to the first, for it also f2 = + 30 cm.
Substituting in the lens formula we get: 1/v2 - (- 1/40) = 1/30 => 1/v2 = 1/30 - 1/40 = 1/120
v2 = +120 cm
The final image is formed 120 cm to the right of the second lens (or 280 cm to the right of the first) and is real and inverted with respect to the image due to the first lens (erect with respect to the object).
c) The second lens is 40 cm to the right of the first. Hence, the image due to the first (object for the second) is 80 cm to the right of the second lens.
Now let us consider the second lens alone with its object placed 80 cm to its right.
In the lens maker's formula, the surface 1 of the lens is the one which is on the object's side. Hence, the surface 2 for the first lens becomes surface 1 for the second and vice versa. Further, now the centres of curvature lie on the same side as the object, hence R1 and R2 are -ve. Substituting R1 = - 30 cm and R2 = - 10 cm in the lens maker's formula we get:
1/f2 = (1.5 - 1)[- 1/30 - (- 1/10] = 0.5[- 1/30 + 1/10] = 1/30
f2 = + 30 cm (The focal length remains unchanged even when the side of the lens on which the object is placed is reversed)
Substituting u2 = - 80 cm and f2 = + 30 cm in the lens formula:
1/v2 - (- 1/80) = 1/30 => 1/v2 = 1/30 - 1/80 = 5/240
v2 = + 48 cm
The final image is formed 48 cm to the left of the second lens (or 8 cm to the left of the first).
d) Substituting u2 = - 80 cm and f = - 40 cm in the lens formula we get:
1/v2 - (- 1/80) = - 1/40 => 1/v2 = - 1/40 - 1/80 = - 12/320
v2 = - 26.67 cm
The final image is formed 26.67 cm to the right of the second lens (or 66.67 cm to the right of the first lens)© BrainMass Inc. brainmass.com December 24, 2021, 10:48 pm ad1c9bdddf>