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Question

# An object 4.0 cm in size, is placed 25.0 cm in front of a concave mirror of a focal length of 15.0 cm. (i) At what distance from the mirror should a screen be placed in order to obtain a sharp image? (ii) Find the size of the image. (iii) Draw a ray diagram to show the formation of the image in this case.

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Solution

## Step 1:Given:Height of the object, ${h}_{1}=4cm$Distance of the object from the mirror, $u=-25cm$The focal length of the mirror, $f=-15cm$Step 2: (i) Finding the distance of the image (v) using the mirror formula:We know that $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$⇒\frac{1}{v}+\frac{1}{-25}=\frac{1}{-15}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{1}{-15}+\frac{1}{25}=\frac{\left(-25+15\right)}{375}=\frac{1}{-37.5}cm\phantom{\rule{0ex}{0ex}}⇒v=–37.5cm$The image is formed object side, 37.5cm from the pole of the concave mirror.Step 3: (ii) Finding the height of the image h2 using the formula:We know that $\frac{{H}_{2}}{{H}_{1}}=\frac{-u}{v}$Where ${H}_{1}$ is the size of the image .Now, $\frac{{H}_{2}}{4cm}=\frac{-\left(-37.5cm\right)}{\left(-25cm\right)}$${H}_{2}=-4×\frac{37.5}{25}=-6cm$∴ size of image = 6cmStep 4: (iii) The ray diagram showing the formation of the image in this case:

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