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In an experiment to measure the focal length of concave mirror it was found that for an object distnace of 0.30 m, the image distance come out to be 0.60 m. Determine focal length with error limits. In an experiment to measure the focal length of concave mirror it was found that for an object distnace of 0.30 m, the image distance come out to be 0.60 m. Determine focal length with error limits.
In an experiment to measure the focal length of concave mirror it was found that for an object distnace of 0.30 m, the image distance come out to be 0.60 m. Determine focal length with error limits.
Case 1: If we take vernier callipers, its least count is : dv = du = 0.01cm = 10-4m Case 2: If we take meter scale, its least count is : dv = du = 0.1cm = 10-3m v-image distance u-object distance f-focal length then 1/v + 1/u = 1/f for v=0.6m and u=0.3m , we have (from above formula) f = 0.2m on differentiating above eqn. for error analysis, we have -dv/v2 -du/u2 = -df/f2 Case 1(vernier scale): df = f2( dv/v2 +du/u2 ) = 0.04*(1/0.36 + 1/0.09)*10-4m = 5.56x10-5 m focal length limits are : ( 0.2 ± 5.56x10-5 ) mts Case 2(metee scale): df = f2( dv/v2 +du/u2 ) = 0.04*(1/0.36 + 1/0.09)*10-3m = 5.56x10-4 m focal length limits are : ( 0.2 ± 5.56x10-4 ) mts --- Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Naga Ramesh IIT Kgp - 2005 batch
Case 1: If we take vernier callipers, its least count is : dv = du = 0.01cm = 10-4m
Case 2: If we take meter scale, its least count is : dv = du = 0.1cm = 10-3m
v-image distance
u-object distance
f-focal length
then 1/v + 1/u = 1/f
for v=0.6m and u=0.3m , we have (from above formula) f = 0.2m
on differentiating above eqn. for error analysis, we have
-dv/v2 -du/u2 = -df/f2
Case 1(vernier scale): df = f2( dv/v2 +du/u2 )
= 0.04*(1/0.36 + 1/0.09)*10-4m
= 5.56x10-5 m
focal length limits are : ( 0.2 ± 5.56x10-5 ) mts
Case 2(metee scale): df = f2( dv/v2 +du/u2 )
= 0.04*(1/0.36 + 1/0.09)*10-3m
= 5.56x10-4 m
focal length limits are : ( 0.2 ± 5.56x10-4 ) mts
---
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Naga Ramesh IIT Kgp - 2005 batch
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