Question

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『An object, 4.0 CM in size, is placed at 25.0 CM in front of concave mirror of focal length 15.0 CM. At what distance from the mirror should a screen be placed in order to obtain a sharp image? find the nature and size of the image.』
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Answer 1

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Given:

• Height of the Object(H₀) = 4 cm.
• Object Distance(u) = -25 cm
• Focal Length = -15 cm

Solution:

Using the Mirror's Formula,

$\bold{⇒\frac{1}{-15} = \frac{1}{-25} + \frac{1}{v} }\\ \\ \bold{⇒ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }\\ \\ ⇒  \bold{v = - 37.5 cm}$

The image is formed behind the mirror also, the Distance is Negative, So, the Image formed is real.

_________________________

As we know that,

$\bold{magnification = \frac{H_1}{H_0}}\\ \\ ⇒ \bold{\frac{H_1}{4} = - (-37.5)/25} \\ \\ ⇒ \bold{H_1 = 37.5 × \frac{4}{25}} \\ \\\bold{⇒ H_1= \frac{150}{25}} \\ \\ ⇒ \bold{H_1= 6 cm}$

_________________________

Since, the Size of the Image is greater than the size of the Object, thus the image is Magnified.

Hence, Characteristics of the Images are:

• Real
• Inverted
• Magnified

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