Question

, an object of height 7 cm is kept at a distance of 25 centimetre in the front of a concave mirror the focal length of the mirror is 15 CM at what distance should be the mirror should a screen be kept to get as a clear image what will be the size and nature of the image

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

Given :

• Object Height (Ho) = 7 cm
• Object Distance (u) = - 25 cm
• Focal Length (f) = - 15 cm

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To Find :

• Image Distance, Size, Nature

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Solution :

We have Mirror Formula :

$\large {\boxed{\sf{\dfrac{1}{f} \: = \: \dfrac{1}{v} \: + \: \dfrac{1}{u}}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{f} \: - \: \dfrac{1}{u}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{-1}{15} \: - \: \dfrac{-1}{25}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{-5 \: + \: 3}{75}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{-2}{75}}} \\ \\ \implies {\sf{v \: = \: - \: 37.5}} \\ \\ {\underline{\sf{\therefore \: Image \: distance \: is \: - \: 37.5 \: cm}}}$

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Formula for magnification is :

$\large {\boxed{\sf{m \: = \: \dfrac{h'}{h_o} \: = \: \dfrac{-v}{u}}}} \\ \\ \implies {\sf{\dfrac{h'}{7} \: = \: - \dfrac{-37.5}{-25}}} \\ \\ \implies {\sf{h' \: = \: \dfrac{-37.5 \: \times \: 7}{25}}} \\ \\ \implies {\sf{h' \: = \: - \: 10.5 \: cm}}$

• Image is Magnified
• Nature of Image is Real and Inverted

Answer 2

Answer:

Given:

Object distance = 25 cm

Focal length = 15 cm

Object height = 7 cm

To find:

Object distance , size and nature of image.

Calculation:

Applying Mirror Formula :

$\boxed{ \sf{ \red{ \large{ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}}}}$

$= > \dfrac{1}{ - 15} = \dfrac{1}{v} + \dfrac{1}{ - 25}$

$= > \dfrac{1}{v} = \dfrac{1}{25} - \dfrac{1}{15}$

$= > \dfrac{1}{v} = \dfrac{6 - 10}{150}$

$= > \dfrac{1}{v} = \dfrac{ - 4}{150}$

$= > \dfrac{1}{v} = \dfrac{ - 1}{37.5}$

$= > v = - 37.5 \: cm$

So image distance = 37.5 cm

Since the sign of the image distance is negative , we can say that the image is real ( as it ia formed before the mirror)

Now ,

$\boxed{ \sf{ \red{ \large{mag = - \dfrac{v}{u}}}}}$

$= > mag. = - \dfrac{( - 37.5)}{-25}$

$= > mag. = - 1.5$

Since sign of magnification is negative , we can say that the image is inverted .

Now , image size :

$\boxed{ \sf{ \large{ \red{ \dfrac{image \: size}{object \: size} = mag.}}}}$

$= > \dfrac{image \: size}{7} = -1.5$

$= > image \: size = -1.5 \times 7$

$= > image \: size = -10.5 \: cm$

So image is also magnified and enlarged.