Question

An object of size 5 cm is placed at a distance of 20 cm in front of a concave mirror focal length 10 cm. The distance of the image from the mirror and its height will be?​

Answer 1

$\bullet \footnotesize\text \purple{ Given : - }$

$\:$

$\purple ➳ \footnotesize\text{ \: u \purple(object \: distance\purple) = - 20 cm}$

$\purple➳ \footnotesize\text{ \: f \purple(focal \: length\purple) = - 10 cm }$

$\purple ➳ \footnotesize\text{ \: ho \purple(height \: of \: object\purple) \: = 5cm }$

$\:$

$\bullet \footnotesize\text \purple{ Solution:- }$

$\:$

$\bullet \footnotesize\text \purple{ Use \: mirror \: formula}$

$\:$

$\purple➳ \tt{ \: \frac{1}{ f} = \frac{1}{v} + \frac{1}{ u} }$

$\purple➳\tt{ \: \frac{1}{ - 10} = \frac{1}{v} + \frac{1}{ - 20} }$

$\purple➳ \tt{ \: \frac{ - 1}{ 10} = \frac{1}{v} - \frac{1}{ 20} }$

$\purple➳\tt{ \: \frac{ - 1}{ 10} + \frac{1}{20} = \frac{1}{v} }$

$\purple➳\tt{ \: \frac{ - 2 + \: 1} {20} = \frac{1}{v} }$

$\purple➳\tt{ \frac{ - 1} {20} = \frac{1}{v} }$

$\purple➳\footnotesize \text{ - v = 20 }$

$\bullet \: \boxed{ \boxed{\footnotesize\text \purple{ v = - 20cm }}}$

$\:$

$\bullet \footnotesize\text \purple{ Now use magnification formula }$

$\:$

$\purple➳\tt{ \:m \purple(magnification \purple) = \frac{ - v}{u} }$

$\purple➳ \tt{ \:m = \frac{ - ( - 20)}{ - 20} }$

$\purple➳\tt{ \:m = \frac{ 20}{ - 20} }$

$\purple➳\tt{ \:m = \cancel\frac{ 20}{ - 20} }$

$\bullet \: \boxed{ \boxed{\footnotesize\text \purple{ m = - 1 }}}$

$\:$

$\purple➳\tt{ \:m \purple(magnification \purple) = \frac{ hi \purple(height \: of \: image \purple)}{ho \purple(height \: of \: object \purple)} }$

$\purple➳\tt{ \: - 1 = \frac{ hi}{5} }$

$\bullet \: \boxed{ \boxed{\footnotesize\text \purple{ hi = - 5cm}}}$

$\:$

✡ Important points:-

✡ In concave mirror u(object distance) and f(focal length) is always negative

✡ In convex mirror u is negative and f is positive

✡ If m value is less than 1 that means image size is diminished

✡ If m = 1 then image is same size as object

✡ If m greater than 1 image size is enlarged

✡ If m is negative image is real and inverted

✡ If m is positive image is virtual and erect

Answer 2

Answer:

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Explanation:

We know that,

v1+u1=f1

u = -10 cm

Focal length = 15 cm

∴v1=151−(−10)1

⇒v1=305=61

⇒v=6cm

∴ The image is formed at a distance 6 cm