A point source S is placed midway between two converging mirrors having equal focal length f as shown in fig. Find the values of d for which only one image is formed.
A
2f, 3f
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3f, 4f
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
f, 2f
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2f, 4f
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D 2f, 4f Here we need only one image, the idea is that both the mirrors are going to form the image of the point object S and the image formed might again serve as an object to find another image and so on. If we need only one image, the ray intersection must occur only at one point always. In other words, both the mirrors must form the image of S at S only. Now, for this, S must be the center of curvature for both the mirrors (As the objects at center of curvature form image at center of curvature only).
So, M1S=M2S=2fWehaved=M1S+M2S=4f.
Also, Let's talk about a special case, when the object is at the common focus, the reflected ray from any of the mirror will be parallel and will focus again after reflection from another mirror. In this case too we will have only one image
In this case, M1S=M2S=fAndwegetd=M1S+M2S=2f.
So, we have two values for d;2f & 4f for only one image to be formed.