An image of the same size as that of the object cannot be produced by:
A. Plane mirror
B. Concave Mirror
C. Convex Mirror
D. Convex lens
Answer
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Hint: Use the magnification formula in each case by substituting value of m=1. Then substitute the values of u and v in the mirror/lens equation.
Complete step by step answer:
Different types of optical lens produce a variety of images. Just by changing the focal length of the same lens/mirror or changing the size of aperture, we can get different types of images. Here, we need to discuss the image produced by the four given optical instruments. So,
Mirror equation: \[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\]; Lens Equation: \[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
Mirror Magnification ${m_m} = - \dfrac{v}{u}$; Lens Magnification: ${m_l} = \dfrac{v}{u}$
Lets first consider the case of mirrors
Convex Mirror: ${m_m} = 1 \Rightarrow v = - u$.
But a convex mirror always forms a virtual erect image so m>0 which means $v = - u$. Substitute this value in the mirror equation we find $f \to \infty $which is not possible. Therefore convex mirrors cannot produce the same size image.
Concave mirror: produces a real inverted image so m < 0; ${m_m} = - 1 \Rightarrow v = u$.
Substitute this value in the mirror equation we find that the concave mirror produces the same size image at u=2f=R.
Plane Mirror: Always produces the same size image.
Convex Lens: produces real inverted images for u < f so m < 0; ${m_l} = - 1 \Rightarrow v = - u$. Substitute this value in the lens equation we find that the convex lens forms the same size image for u > f at u=-2f.
For u
So, the correct answer is “Option C”.
Note:
Some lenses for a given position of an object just change the whole nature of the image like in the case of convex lenses. If it would have been given a convex lens for object distance less than focal length then the answer would be both convex mirror and convex lens for u < f.
Complete step by step answer:
Different types of optical lens produce a variety of images. Just by changing the focal length of the same lens/mirror or changing the size of aperture, we can get different types of images. Here, we need to discuss the image produced by the four given optical instruments. So,
Mirror equation: \[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\]; Lens Equation: \[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
Mirror Magnification ${m_m} = - \dfrac{v}{u}$; Lens Magnification: ${m_l} = \dfrac{v}{u}$
Lets first consider the case of mirrors
Convex Mirror: ${m_m} = 1 \Rightarrow v = - u$.
But a convex mirror always forms a virtual erect image so m>0 which means $v = - u$. Substitute this value in the mirror equation we find $f \to \infty $which is not possible. Therefore convex mirrors cannot produce the same size image.
Concave mirror: produces a real inverted image so m < 0; ${m_m} = - 1 \Rightarrow v = u$.
Substitute this value in the mirror equation we find that the concave mirror produces the same size image at u=2f=R.
Plane Mirror: Always produces the same size image.
Convex Lens: produces real inverted images for u < f so m < 0; ${m_l} = - 1 \Rightarrow v = - u$. Substitute this value in the lens equation we find that the convex lens forms the same size image for u > f at u=-2f.
For u
So, the correct answer is “Option C”.
Note:
Some lenses for a given position of an object just change the whole nature of the image like in the case of convex lenses. If it would have been given a convex lens for object distance less than focal length then the answer would be both convex mirror and convex lens for u < f.
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