In an optical bench experiment index error is \[ + 1\,{\text{cm}}\] and \[ - 1\,{\text{cm}}\], between object needle and lens, lens and image needle respectively. Observed values of u and v are \[9\,{\text{cm}}\]and \[17\,{\text{cm}}\]. Focal length of lens is closest to:-
A. \[{\text{6}}{\text{.15 cm}}\]
B. \[{\text{5}}{\text{.54 cm}}\]
C. \[{\text{5}}{\text{.88 cm}}\]
D. \[{\text{6}}{\text{.25 cm}}\]
Answer
593.7k+ views
Hint: For this question, use the lens formula to find out the focal length of the lens.
In the lens formula put the values of u and v to get the value of focal length of lens.
Complete step by step answer:
Given, distance between needle and the lens, \[u = - {\text{9cm}}\]
(the negative sign is because of the sign convention)
Distance between image and the lens, \[v = 17{\text{cm}}\]
From the lens formula we have,
\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\]...............................(1)
Where \[u\] is the distance between object and the lens and \[v\] is the distance between image and the lens and \[f\] is the focal length of the lens.
Putting the values of \[u\] and \[v\] in equation (1), we get
\[\dfrac{1}{{17}} - \dfrac{1}{{\left( { - 9} \right)}} = \dfrac{1}{f}\]
\[ \Rightarrow \dfrac{1}{f} = \dfrac{{26}}{{153}}\]
\[ \Rightarrow f = \dfrac{{153}}{{26}} = 5.88\,{\text{cm}}\]
Therefore, focal length of the lens is closest to \[5.88\,{\text{cm}}\].
So, the correct answer is “Option C”.
Note:
While solving such questions in which we have to use lens or mirror formulas take care of the sign convention, the distance from the left of the lens or mirror are taken negative and distances taken from the right are taken as positive.
Optics Bench provides the basic knowledge about properties of light such as Refraction, Reflection, Dispersion and Total internal reflection.
In the lens formula put the values of u and v to get the value of focal length of lens.
Complete step by step answer:
Given, distance between needle and the lens, \[u = - {\text{9cm}}\]
(the negative sign is because of the sign convention)
Distance between image and the lens, \[v = 17{\text{cm}}\]
From the lens formula we have,
\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\]...............................(1)
Where \[u\] is the distance between object and the lens and \[v\] is the distance between image and the lens and \[f\] is the focal length of the lens.
Putting the values of \[u\] and \[v\] in equation (1), we get
\[\dfrac{1}{{17}} - \dfrac{1}{{\left( { - 9} \right)}} = \dfrac{1}{f}\]
\[ \Rightarrow \dfrac{1}{f} = \dfrac{{26}}{{153}}\]
\[ \Rightarrow f = \dfrac{{153}}{{26}} = 5.88\,{\text{cm}}\]
Therefore, focal length of the lens is closest to \[5.88\,{\text{cm}}\].
So, the correct answer is “Option C”.
Note:
While solving such questions in which we have to use lens or mirror formulas take care of the sign convention, the distance from the left of the lens or mirror are taken negative and distances taken from the right are taken as positive.
Optics Bench provides the basic knowledge about properties of light such as Refraction, Reflection, Dispersion and Total internal reflection.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
Which is more stable and why class 12 chemistry CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What are the major means of transport Explain each class 12 social science CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

