Question

15. An object of size 7.0 cm is placed at 27 cm 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 the nature of the image.



in solution I see that u=-27 and f=-18
so how - is coming

Answer 1

Answer:

given-

u=-27cm

h(o)=7cm

f=-18cm

soln-

using mirror formula,

=1/f=1/u+1/v

=1/-18=1/-27+1/v

=1/v=1/27-1/18

=1/v=2-3/54

=1/v=-1/54

therefore. v=-54

so screen should ne placed at 54cm in front Of mirror

here,

m=-v/u=h(i)/h(o)

=-(-54)/-27

= 54/-27

= -2

wo,m is greater than 1 and -ve

so nature Of image=real and inverted

size=large from object size

theref

Explanation:

you see u and f is negative because according to sign convention distances left to mirror is negative .

so f and u is -ve because it is situated left to the mirror..

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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