An object 2cm in size is placed 30cm in front of a concave mirror of focal length 15cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? What will be the nature and the size of the image formed? Draw a ray diagram to show the formation of the image in this case.
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$\text{u}=-30$$\text{f}=-15$
$\frac{1}{\text{u}}+\frac{1}{\text{v}}=\frac{1}{\text{f}}$
$\frac{-1}{30}+\frac{1}{\text{v}}=\frac{-1}{15}$
$\frac{1}{\text{v}}=\frac{-1}{15}+\frac{1}{30}$
$\frac{1}{\text{v}}=\frac{-1}{30}$
${\text{v}}=-30$
Therefore the screen must be placed 30cm in front of the mirror
Now, the object is kept at 30cm which is equal to twice the focal length that is 2f
Also r that is distance between centre and pole = 2f
Therefore the object is kept at the centre so the image formed will also be of 2cm, real and inverted.
$\frac{1}{\text{u}}+\frac{1}{\text{v}}=\frac{1}{\text{f}}$
$\frac{-1}{30}+\frac{1}{\text{v}}=\frac{-1}{15}$
$\frac{1}{\text{v}}=\frac{-1}{15}+\frac{1}{30}$
$\frac{1}{\text{v}}=\frac{-1}{30}$
${\text{v}}=-30$
Therefore the screen must be placed 30cm in front of the mirror
Now, the object is kept at 30cm which is equal to twice the focal length that is 2f
Also r that is distance between centre and pole = 2f
Therefore the object is kept at the centre so the image formed will also be of 2cm, real and inverted.
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