The size of the image formed by a convex mirror is half the size of the object, when the latter is held at a distance of 30 cm from the mirror. Find the radius of curvature of the mirror.
Answer
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Hint:Determine the image distance using the formula for magnification of the lens. Using the lens formula, determine the focal length of the convex lens. It should be a positive. The radius of curvature is twice the focal length of the lens.
Formula used:
Magnification, \[m = \dfrac{{{h_i}}}{{{h_o}}}\]
Here, \[{h_i}\] is the height of the image and \[{h_o}\] is the height of the object.
Lens formula, \[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\]
Here, f is the focal length, u is the object distance and v is the image distance.
Complete step by step answer:
We have given that the size of the image is half the size of the object. We have the expression for the magnification of the lens,
\[m = \dfrac{{{h_i}}}{{{h_o}}}\]
Here, \[{h_i}\] is the height of the image and \[{h_o}\] is the height of the object.
According to the question,
\[\dfrac{{{h_i}}}{{{h_o}}} = \dfrac{1}{2} = m\] …… (1)
We also have the expression for the magnification in terms of image distance and object distance,
\[m = - \dfrac{v}{u}\]
Here, v is the image distance and u is the object distance.
Using equation (1) in the above equation, we get,
\[\dfrac{1}{2} = - \dfrac{v}{u}\]
\[ \Rightarrow \dfrac{1}{2} = - \dfrac{v}{{\left( { - 30} \right)}}\]
\[ \Rightarrow v = 15\,{\text{cm}}\]
We have the lens equation,
\[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\]
Substituting \[v = 15\,{\text{cm}}\] and \[u = - 30\,{\text{cm}}\] in the above equation, we get,
\[\dfrac{1}{f} = \dfrac{1}{{15}} + \dfrac{1}{{ - 30}}\]
\[ \Rightarrow \dfrac{1}{f} = \dfrac{2}{{30}} - \dfrac{1}{{30}}\]
\[ \Rightarrow f = 30\,{\text{cm}}\]
We know that the radius of curvature of the lens is twice the focal length. Therefore,
\[R = 2f\]
Substituting \[f = 30\,{\text{cm}}\] in the above equation, we get,
\[R = 2\left( {30} \right)\]
\[ \therefore R = 60\,{\text{cm}}\]
Therefore, the radius of curvature of the convex lens is 60 cm.
Note: Students must be able to recognize the nature of the lens using the sign of focal length. The convex lens has a positive focal length while the concave lens has a negative focal length. Also, the object distance is always negative for both convex lens and concave lens while the image distance is positive for real image formed by convex lens.
Formula used:
Magnification, \[m = \dfrac{{{h_i}}}{{{h_o}}}\]
Here, \[{h_i}\] is the height of the image and \[{h_o}\] is the height of the object.
Lens formula, \[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\]
Here, f is the focal length, u is the object distance and v is the image distance.
Complete step by step answer:
We have given that the size of the image is half the size of the object. We have the expression for the magnification of the lens,
\[m = \dfrac{{{h_i}}}{{{h_o}}}\]
Here, \[{h_i}\] is the height of the image and \[{h_o}\] is the height of the object.
According to the question,
\[\dfrac{{{h_i}}}{{{h_o}}} = \dfrac{1}{2} = m\] …… (1)
We also have the expression for the magnification in terms of image distance and object distance,
\[m = - \dfrac{v}{u}\]
Here, v is the image distance and u is the object distance.
Using equation (1) in the above equation, we get,
\[\dfrac{1}{2} = - \dfrac{v}{u}\]
\[ \Rightarrow \dfrac{1}{2} = - \dfrac{v}{{\left( { - 30} \right)}}\]
\[ \Rightarrow v = 15\,{\text{cm}}\]
We have the lens equation,
\[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\]
Substituting \[v = 15\,{\text{cm}}\] and \[u = - 30\,{\text{cm}}\] in the above equation, we get,
\[\dfrac{1}{f} = \dfrac{1}{{15}} + \dfrac{1}{{ - 30}}\]
\[ \Rightarrow \dfrac{1}{f} = \dfrac{2}{{30}} - \dfrac{1}{{30}}\]
\[ \Rightarrow f = 30\,{\text{cm}}\]
We know that the radius of curvature of the lens is twice the focal length. Therefore,
\[R = 2f\]
Substituting \[f = 30\,{\text{cm}}\] in the above equation, we get,
\[R = 2\left( {30} \right)\]
\[ \therefore R = 60\,{\text{cm}}\]
Therefore, the radius of curvature of the convex lens is 60 cm.
Note: Students must be able to recognize the nature of the lens using the sign of focal length. The convex lens has a positive focal length while the concave lens has a negative focal length. Also, the object distance is always negative for both convex lens and concave lens while the image distance is positive for real image formed by convex lens.
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