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Home / Doubts and Solutions/JEE/Physics

A concave mirror,a screen and object are arranged such that the height of image is doubled that of the object.then the screen and object movies to get a triple magnified image.in this case screen is moved by25cm.find the focal length of mirror?
 

Asked by venkat 31st July 2017, 2:37 PM
Answered by Expert
Answer:
The following steps are needed to solve the query:
 
1) Take the magnification (real image, so m = -2) for first set of object and image distances u1 and v1. Using that, you will get u1 = v1/2.
 
2) Use the mirror formula to obtain a relation between f and v1. You will get f = v1/3.
 
3) Now, take the magnification formula for the second set of object and image distances u2 and v2. You will get u2 = v2/3.
 
4) Use the mirror formula to obtain a relation between f and v2. You will get f = v2/4.
 
5) Use the concept of image formation by a concave mirror. The concept tells us that a concave mirror forms a magnified real image with increasing magnification when the distance of the object is reduced and the image distance is increased. So, now we know that when the screen is moved by 25 cm, it indicates that v2 = v1 + 25.
 
6) Substitute value of v2 in step 4 to get a relation between f and v1 again.
 
7) Equate both the values of f as the focal length of the mirror remains the same. You will get v1/3 = (v1 + 25)/4. Solving this you will get v1 = 75 cm.
 
8) Use f = v1/3 to obtain f = 25 cm.
 
NOTE: Once you substitute actual values of u, you can use sign convention and you will obtain that f = -25 cm. But, without values this is the final answer.
Answered by Expert 1st August 2017, 1:41 PM
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