Question

A 9.35 mm high chocolate chip is placed on the axis of, and 12.9 cm from, a lens with a focal length of 5.65 cm. If it can be determined, is the chocolate chip's image real or virtual? cannot be determined O virtual Oreal How high is the image? Express the answer as a positive quantity. image height: mm If it can be determined, is the image upright or inverted with respect to the real thing? upright inverted O cannot be determined

Answer 1

Assuming that the lens used here is convex lens.

If focal length is f= 5.65 cm, then centre of curvature, R = 2f = 11.3 cm

The Object, chocolate chip at a distance of 12.9 cm

u = -12.9 cm (by convention, lengths on left of lens is taken as negative and right as positive)

Ray Diagram when object is placed beyond R

Therefore, the object lies beyond R, and according to ray diagram, we know that the image will be formed between f and 2R on right side of lens, and the image will be real and inverted.

For convex lens, f is taken to be positive.

Using Lens Formula

11 1 - - - =-

1/v = 1/f + 1/u = 1/5.65 + 1/(-12.9) = 0.178 - 0.078 = 0.099

v = 1/0.099 = 10.05 cm , this is positive, that means, image formed is on the right of the lens ( and notice that it lies between f and R as we predicted)

Image height can be determined from magnification formula given by

height of image (h') m = mehr m height of object (h)

m = v/u = 10.05/(-12.9) = -0.78

Now, according to convention, objects/images above principle axis are taken to be positive and the ones below are negative.

So, m = height of image/ height of object.

-0.78 = h/ 9.35

h = -7.29 mm , that is, the image is below principle axis, hence inverted.

Hence, image is REAL, INVERTED and of Size 7.29 mm