## Friday, January 9, 2015

### Problems and Solutions on Refraction of Light

Problem and solution

A 2 cm high object is placed on the principal axis of a concave mirror at a distance of 12 cm from the pole. If the image is inverted, real and 5 cm in height, find the location of the image and the focal length of the mirror?

We can solve this problem basing on the very definition of magnification and the mirror formula. Magnification is defined as the ratio of height of the image to the height of the object. The other way of defining the magnification is as the ratio of distance of the image to the distance of the object. The sign of the magnification is negative which means that object and image are in the different directions. The problem is solved as shown below.

An object is placed in the principal axis of a concave mirror at a distance X  from a principal focus. The images formed at a distance Y the focus. What is the focal length of the mirror ?

This problem also has to be solved basing on the mirror formula. Being both object and the major place to before the mirror they shall be treated as negative and a concave mirror focal length is also negative.The solution is shown in the above diagram.

Problem and solution

An object is placed in front of a concave mirror at a distance of 50 cm. A plain mirror is introduced covering the lower half of a convex mirror. The distance between the object and the plain mirror is 30 cm. It is found that there is no gap between the image formed by the two mirror. Then what is the radius of curvature of the convex mirror?

In the problem object is placed at a distance of 50 cm from a concave mirror. Between the object and the mirror at a distance 30 cm from the object a plain mirror is placed. That means the plain mirror is a distance of 20 cm from the convex mirror. The image of the object due to the plain mirror will be farmed again at the 30 cm as a virtual image in the backward direction as shown. And hence it is going to be 10 cm behind the convex mirror.

The same shall be the image of the convex mirror also as it is given in the problem that both the images are coinciding with each other.

Therefore we know that the object is in front of the mirror at a distance of 50 cm and the image is behind the concave mirror at a distance of 10 cm and using the mirror formula we can calculate the focal as shown below.

Problem and solution

Two blocks each of mass m lies on a smooth table. They are attracted to the two other masses as shown. The pullies are straight and light. An object to is kept on the table as shown. The surfaces of the two blocks are made are reflecting surfaces. Find the acceleration of the two images by the two reflecting surfaces with respect to each other?

We shall try equations of motion using Newton’s laws. We shall draw  free body diagram and identify the direction of motion. The forces along the direction of motion shall be treated as positive and vice versa. As shown in the below diagram, we can write the equations of motion and derive acceleration of the individual images.

As per the law of optics, the acceleration of the image is twice the acceleration of the mirror.

As the two images are moving in the opposite direction, we can calculate the relative acceleration is the sum of the two accelerations.

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