Physics, asked by suprabhachoudhury143, 8 months ago

Sharp focussed amage can be obtained a find the size
Q4 An object of size 7.0 cm & placed at 27 cm in front of
la Concare mirtor of focal length 18cm. At what distance
from the mirror should a screen be placed, so that a sharp focussed image can be obtained?find the size
& the nature of the image can be obtained? find the size and the nature of the image

?​

Answers

Answered by prajwaldeshmukh350
1

Answer:

When object is between C and F

The image is beyond C / between infinite and C

is Real and inverted

is enlarged

Magnification (m)=-ve and |m|>1

Attachments:
Answered by Anonymous
4

•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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