
An object is placed at a distance of$30cm$from a concave mirror of focal length $15cm$. List four characteristics (nature, position etc) of the image formed by the lens.
Answer
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Hint: The concave mirror can form both real and virtual images for different positions of the object. The mirror formula gives us the relation between object distance, image distance and focal length. By substituting the corresponding values, we can find the missing value. The ratio of size of image to the size of object is called magnification.
Formula used:
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
$M=-\dfrac{v}{u}$
Complete answer:
A concave mirror is a mirror curved inwards which can form real as well as virtual images for different positions of an object. By convention, the object distance and focal length are negative.
The mirror formula is given as-
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
Here,$f$is the focal length of the mirror
$v$is the image distance
$u$is the object distance
Given $u=-30cm,\,f=-15cm$
In the above equation, we substitute the given values to get,
$\begin{align}
& \dfrac{1}{-15}=\dfrac{1}{v}+\dfrac{1}{-30} \\
& \Rightarrow \dfrac{-1}{15}+\dfrac{1}{30}=\dfrac{1}{v} \\
& -\dfrac{1}{30}=\dfrac{1}{v} \\
& \therefore v=-30cm \\
\end{align}$
Therefore, the image is formed at a distance of $30cm$ in front of the mirror, at the same side as the object. The image formed is real and inverted.
The formula for magnification is-
$M=-\dfrac{v}{u}$
Substituting values in the above equation, we get,
$\begin{align}
& M=-\dfrac{-30}{-30} \\
& \therefore M=-1 \\
\end{align}$
As the magnification is unity, the image is of the same size as the object. As magnification is negative, the image is inverted with respect to the object.
Therefore, the image is formed at $30cm$ in front of the mirror. It is a real and inverted image and its size is equal to the object.
Note:
The image is formed at the centre of curvature which is at a distance equal to radius of curvature from the centre of the mirror. For all distances beyond focus, a concave mirror forms real and inverted images. When the object is kept between the centre of mirror and focus, the images formed are virtual and erect.
Formula used:
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
$M=-\dfrac{v}{u}$
Complete answer:
A concave mirror is a mirror curved inwards which can form real as well as virtual images for different positions of an object. By convention, the object distance and focal length are negative.
The mirror formula is given as-
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
Here,$f$is the focal length of the mirror
$v$is the image distance
$u$is the object distance
Given $u=-30cm,\,f=-15cm$
In the above equation, we substitute the given values to get,
$\begin{align}
& \dfrac{1}{-15}=\dfrac{1}{v}+\dfrac{1}{-30} \\
& \Rightarrow \dfrac{-1}{15}+\dfrac{1}{30}=\dfrac{1}{v} \\
& -\dfrac{1}{30}=\dfrac{1}{v} \\
& \therefore v=-30cm \\
\end{align}$
Therefore, the image is formed at a distance of $30cm$ in front of the mirror, at the same side as the object. The image formed is real and inverted.
The formula for magnification is-
$M=-\dfrac{v}{u}$
Substituting values in the above equation, we get,
$\begin{align}
& M=-\dfrac{-30}{-30} \\
& \therefore M=-1 \\
\end{align}$
As the magnification is unity, the image is of the same size as the object. As magnification is negative, the image is inverted with respect to the object.
Therefore, the image is formed at $30cm$ in front of the mirror. It is a real and inverted image and its size is equal to the object.
Note:
The image is formed at the centre of curvature which is at a distance equal to radius of curvature from the centre of the mirror. For all distances beyond focus, a concave mirror forms real and inverted images. When the object is kept between the centre of mirror and focus, the images formed are virtual and erect.
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