
A concave mirror of focal length 10 cm is placed at a distance of 35 cm from a wall. How far from the wall an object be placed so that its image formed by the mirror falls on the wall.
Answer
508.2k+ views
Hint: The mirror formula should be used to solve this problem. The mirror formula gives the relation between the focal length, the object distance and the image distance. The values of the focal length and the image distance are given, using these values, we will compute the object distance. By subtracting the object distance from the image distance, we will get the required result.
Formula used:
\[\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}\]
Complete step-by-step solution:
From the given information, we have the data as follows.
A concave mirror of focal length 10 cm is placed at a distance of 35 cm from a wall.
The focal length, \[f=-10\,cm\]
The image distance, \[v=-35\,cm\]
Let ‘x’ be the distance between the object and the wall.
The object distance from the wall, \[u=-(35-x)\]
The mirror formula gives the relation between the focal length, the object distance and the image distance. The mathematical representation of the same is given as follows.
\[\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}\]
The values of the focal length and the image distance are given, using these values, we will compute the object distance and the distance of the object from the wall.
Substitute the values in the above equation.
\[\begin{align}
& \dfrac{1}{-10}=\dfrac{1}{-35}+\dfrac{1}{-(35-x)} \\
& \Rightarrow \dfrac{1}{(35-x)}=\dfrac{1}{14} \\
& \Rightarrow 35-x=14 \\
& \Rightarrow x=35-14 \\
& \therefore x=21\,cm \\
\end{align}\]
\[\therefore \] From the distance of 21 cm, an object must be placed from the wall, so that its image formed by the mirror falls on the wall.
Note: The values of the focal length and the image distance should be considered to be negative. The mirror formula gives the relation between the focal length, the object distance and the image distance. The sign (negative/positive) is assigned to the values of the focal length, image distance and the object distance by taking care of the position of the object concerning the lens/mirror.
Formula used:
\[\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}\]
Complete step-by-step solution:
From the given information, we have the data as follows.
A concave mirror of focal length 10 cm is placed at a distance of 35 cm from a wall.
The focal length, \[f=-10\,cm\]
The image distance, \[v=-35\,cm\]
Let ‘x’ be the distance between the object and the wall.
The object distance from the wall, \[u=-(35-x)\]
The mirror formula gives the relation between the focal length, the object distance and the image distance. The mathematical representation of the same is given as follows.
\[\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}\]
The values of the focal length and the image distance are given, using these values, we will compute the object distance and the distance of the object from the wall.
Substitute the values in the above equation.
\[\begin{align}
& \dfrac{1}{-10}=\dfrac{1}{-35}+\dfrac{1}{-(35-x)} \\
& \Rightarrow \dfrac{1}{(35-x)}=\dfrac{1}{14} \\
& \Rightarrow 35-x=14 \\
& \Rightarrow x=35-14 \\
& \therefore x=21\,cm \\
\end{align}\]
\[\therefore \] From the distance of 21 cm, an object must be placed from the wall, so that its image formed by the mirror falls on the wall.
Note: The values of the focal length and the image distance should be considered to be negative. The mirror formula gives the relation between the focal length, the object distance and the image distance. The sign (negative/positive) is assigned to the values of the focal length, image distance and the object distance by taking care of the position of the object concerning the lens/mirror.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

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

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

