An object is 24 cm away from a concave mirror and its image is 16 cm from the mirror. Find the focal length and radius of curvature of the mirror, and the magnification of the image.
Answer
650.1k+ views
Hint: Use the lens formula to determine the focal length of the mirror. The image formed by the concave mirror is on the same side of the lens as the object. The radius of curvature of the mirror or lens is twice the focal length of the mirror.
Formula used:
\[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\]
Here, f is the focal length, v is the object distance and u is the image distance.
Complete step by step answer:For a concave mirror, the focal length is negative. Also, the object and image are on the same side of the mirror. therefore, the distance of both object and image is negative.
Use lens formula to determine the focal length of the concave mirror as follows,
\[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\]
Here, f is the focal length, v is the object distance and u is the image distance.
Substitute \[ - 16\,cm\] for v and \[ - 24\,cm\] for u in the above equation.
\[\dfrac{1}{f} = \dfrac{1}{{ - 16}} + \dfrac{1}{{ - 24}}\]
\[ \Rightarrow \dfrac{1}{f} = \dfrac{{ - \left( {24 + 16} \right)}}{{384}}\]
\[ \Rightarrow \dfrac{1}{f} = - 0.104\]
\[\therefore f = - 9.6\,cm\]
Therefore, the focal length of the given concave mirror is \[ - 9.6\,cm\].
The relation between the radius of curvature and focal length is,
\[R = 2f\]
Substitute \[ - 9.6\,cm\] for f in the above equation.
\[R = 2\left( { - 9.6\,cm} \right)\]
\[\therefore R = - 19.2\,cm\]
Therefore, the radius of the curvature of the given mirror is \[ - 19.2\,cm\].
The formula for the magnification of the mirror is,
\[m = - \dfrac{v}{u}\]
Substitute \[ - 16\,cm\] for v and \[ - 24\,cm\] for u in the above equation.
\[m = - \dfrac{{ - 16\,cm}}{{ - 24\,cm}}\]
\[m = - \dfrac{2}{3}\]
Therefore, the magnification of the given concave mirror is \[ - \dfrac{2}{3}\].
Note:Always choose the positive and negative scale for the object distance and image distance. For simplicity, take the distance towards the left from the mirror as negative and the distance towards the right of the lens as positive.
Formula used:
\[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\]
Here, f is the focal length, v is the object distance and u is the image distance.
Complete step by step answer:For a concave mirror, the focal length is negative. Also, the object and image are on the same side of the mirror. therefore, the distance of both object and image is negative.
Use lens formula to determine the focal length of the concave mirror as follows,
\[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\]
Here, f is the focal length, v is the object distance and u is the image distance.
Substitute \[ - 16\,cm\] for v and \[ - 24\,cm\] for u in the above equation.
\[\dfrac{1}{f} = \dfrac{1}{{ - 16}} + \dfrac{1}{{ - 24}}\]
\[ \Rightarrow \dfrac{1}{f} = \dfrac{{ - \left( {24 + 16} \right)}}{{384}}\]
\[ \Rightarrow \dfrac{1}{f} = - 0.104\]
\[\therefore f = - 9.6\,cm\]
Therefore, the focal length of the given concave mirror is \[ - 9.6\,cm\].
The relation between the radius of curvature and focal length is,
\[R = 2f\]
Substitute \[ - 9.6\,cm\] for f in the above equation.
\[R = 2\left( { - 9.6\,cm} \right)\]
\[\therefore R = - 19.2\,cm\]
Therefore, the radius of the curvature of the given mirror is \[ - 19.2\,cm\].
The formula for the magnification of the mirror is,
\[m = - \dfrac{v}{u}\]
Substitute \[ - 16\,cm\] for v and \[ - 24\,cm\] for u in the above equation.
\[m = - \dfrac{{ - 16\,cm}}{{ - 24\,cm}}\]
\[m = - \dfrac{2}{3}\]
Therefore, the magnification of the given concave mirror is \[ - \dfrac{2}{3}\].
Note:Always choose the positive and negative scale for the object distance and image distance. For simplicity, take the distance towards the left from the mirror as negative and the distance towards the right of the lens as positive.
Recently Updated Pages
What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

A Paragraph on Pollution in about 100-150 Words

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

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

