JEE Mains · Physics · STD 12 - Ray optics and optical instruments
- JEE Mains 2023
- Hard
- Numerical
In an experiment for estimating the value of focal length of converging mirror, image of an object placed at \(40\,cm\) from the pole of the mirror is formed at distance \(120\,cm\) from the pole of the mirror. These distances are measured with a modified scale in which there are \(20\) small divisions in \(1\,cm\). The value of error in measurement of focal length of the mirror is \(1 / K\) \(cm\). The value of \(K\) is \(..........\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(32\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{ v }+\frac{1}{ u }=\frac{1}{ f }\)
\(\frac{-1}{120}-\frac{1}{40}=\frac{1}{ f }, \quad f =-30\,cm\)
Now,
\(\frac{-1}{ v ^2} dv -\frac{1}{ u ^2} du =-\frac{1}{ f ^2} df\)
Also \(dv = du =\frac{1}{20} cm\)
\(\frac{\frac{1}{20}}{(120)^2}+\frac{\frac{1}{20}}{(40)^2}=\frac{d f}{(30)^2}\)
On solving
\(df =\frac{1}{32}\,cm\)
\(k =32\)
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