Question

An object of size 7.0 cm is placed at 27cm in front of a concave mirror of focal length 18 cm . at what distance from the mirror should a screen be placed,so that a sharp focussed image can be obtained ? find the size and nature of the image.

Answer 1

U=-27,F=-18,Ho=7.o then by mirror formula1/v+1/u=1/f so 1/v+(-1/27)=-1/18then v=-54, but m=v/u=54/27=2,also,m=hi/ho, so,2=hi/7.0now, hi=14.0,hence image image is real and inlarge

Answer 2

•GIVEN:-

$\bf\:➠ Object \: distance \: ,u=-27cm$

$\bf {➠Object \: height,h=7cm}$

$\bf ➠Focal \: length=-18cm$

$\bf \large\underline{\overline{FORMULA \: USED⇓⇓}}$

$\bf \boxed{➣ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }$

SOLUTION :-

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

$\frac{1}{v} = \bf \: \frac{1}{( - 18)} - \frac{1}{ (- 27)}$

$\frac{1}{v} = \bf \: \frac{1}{( - 18)} + \frac{1}{27}$

$\frac{1}{v} = \frac{( - 3 + 2)}{54} = \frac{ - 1 \: \: }{54 }$

$\boxed { \therefore \: v = \bf \: 54cm }$

•The screen should be placed at a distance of 54cm in front of the given mirror

$\bf \: ➽Magnification(m) = \frac{Image \: distance}{Object \: distance}$

$\bf \: m = \frac{ - 54}{ - 27} = - 2$

•Negative value of magnification shows that image is real.

$\bf \: m = \frac{Height \: of \: image}{Height \: of \: object} = \frac{h_1}{h}$

$\bf ➠h_1=m×h=(-2)×7=-14cm \\ \\ </p><p></p><p>$

➽The height of the image is negative which shows that image is inverted.

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