Question

an object 5 cm in length is placed at a distance of 20cm in front of concave mirror of radius of curvature 30cm.Find the position of the image‚ its nature and size​

Answer 1

Given:

• Height of object, $\sf h_o$ = 5 cm
• Object distance, u = - 20 cm
• Radius of curvature, R = 30 cm
• Focal length, f = R/2 = 30/2 = - 15 cm

To find:

• Position of image, it's nature and size?

Solution:

$\dag\;{\underline{\frak{Using\;mirror\;formula\;:}}}$

$\star\;{\boxed{\sf{\purple{ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}}}}$

$:\implies\sf \dfrac{1}{- 15} = \dfrac{1}{v} + \dfrac{1}{- 20}$

$:\implies\sf \dfrac{1}{v} = \dfrac{1}{- 15} - \dfrac{1}{ - 20}$

$:\implies\sf \dfrac{1}{v} = \dfrac{1}{- 15} + \dfrac{1}{20}$

$:\implies\sf \dfrac{1}{v} = \dfrac{- 4 + 3}{60}$

$:\implies\sf \dfrac{1}{v} = \dfrac{- 1}{60}$

$:\implies{\boxed{\frak{\pink{v = - 60\;cm}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Image\;distance\;is\; \bf{- 60\;cm}.}}}$

★ Nature of image:

• Image is virtual and erect.
• Image formed behind the mirror.

⠀━━━━━━━━━━━━━━━━━━━━━━━━━

Size of image,

$\dag\;{\underline{\frak{Using\; Formula\;of\;magnification\;:}}}$

$\star\;{\boxed{\sf{\purple{ \dfrac{h_i}{h_o} = \dfrac{v}{u}}}}}$

$:\implies\sf \dfrac{h_i}{5} = \cancel{ \dfrac{- 60}{20}}$

$:\implies\sf h_i = - 3 \times 5$

$:\implies{\boxed{\frak{\pink{h_i = - 15\;cm}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Height\;or\;size\;of\;image\;is\; \bf{- 15\;cm}.}}}$

Answer 2

Heya !

Given:-

• Height of the object = 5cm.
• Object Distance (u) = -20cm.
• Radius of curvature (r) = -30cm.
• so, f = -30/2 = -15

To find:-

• Position, Size and Nature of the image.

Formula to be used :-

$\huge \fbox \blue{1/v-1/u=1/f}$

Solution:-

$\implies$ 1/v = 1/f+1/u

$\implies$ 1/v = 1/-15+1/-20

$\implies$ 1/v = -4+3 /60

$\implies$ 1/v = -1/60

$\implies$ v = -60 cm.

• the image is at a dist. of 60cm. in front of the mirror.
• this shows that the image is beyond C.
• it's -ve sign shows that it is real and inverted.

Size of the image:-

Formula to be used here:-

$\huge \fbox \blue{m= -v/u}$

Where,

• m = magnification
• v = image Distance
• u = object distance

$\implies$ m = -(-60)/-30

$\implies$ m = -2

when, m is greater than 1 it shows that the image is greater than the object.

• Hence, the image is enlarged as well.