
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
509.1k+ 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.
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