Question

cm
solue the following examples
In object of height 7 cm is kept
kept at a
distance of 25 in front of a
mirror. The focal length of
is 15 cm. At what distance
from the mirror should a screen be kept
as to
image
? what will
and nature of the image?
concave
the mirror is
so
get a clear
be the size
Solution​

Answer 1

Given:-

• Height of object ,ho = 7 cm

• object distance ,u = -25 cm

• Focal length ,f = -15 cm

To Find :-

• Image distance , Nature of Image and Size of image .

Solution:-

Firstly we calculate the Image distance .

Using mirror formula .

• 1/f = 1/v + 1/u

where ,

• f denote focal length
• v denote image distance
• u denote object distance

Substitute the value we get

→ 1/(-15) = 1/v + 1/(-25)

→ -1/15 = 1/v - 1/25

→ 1/v = -1/15 + 1/25

→ 1/v = -5+3 /75

→ 1/v = -2/75

→ v = -75/2

→ v = -37.5

• Therefore, the image distance is 37.5 cm negative sign show that the image is formed as the same side where the object is placed.

Now,

Magnification

• m = -v/u = hi/ho

substitute the value we get

→ -37.5/-25 = hi/7

→ 1.48 = hi/7

→ hi = 7×1.48

→ hi = 10.36

• Height of image is 10.36 cm . Image height is greater than object height.

Nature of Image :- The image formed is Real and Inverted .

Answer 2

Answer:

Given :-

• Height of object = 7 cm
• Focal length = 15 cm
• Object distance = -25 cm

To Find :-

• Nature, Size and distance

Solution :-

As we know that

$\huge \bf \: \frac{1}{u} + \frac{1}{v} = \frac{1}{f}$

U denote object distance

V denote Image distance

F denote Focal length

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

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

$\sf \: \dfrac{1}{v} = \dfrac{ - 5 + 3}{75}$

$\sf \: \dfrac{1}{v} = \dfrac{ - 2}{75}$

$\sf \: v = \dfrac{ - 75}{2}$

$\sf \: v = - 32.5$

Therefore, the image distance is 37.5 cm .

$\huge \bf \: m = \frac{ - v}{u} = \frac{hi}{ho}$

$\sf \: \dfrac{-37.5}{-25}= \dfrac{hi}{7}$

$\sf \: 1.48 = \dfrac{hi}{7}$

$\sf \: hi \: = 7×1.48$

$\sf \: hi \: = 10.36$

Nature :- The image formed is Real and Inverted .