Question

if an object 4 cm height is placed at a distance of 18 cm from concave mirror having focal length 12 cm.find the position, nature, and height of images​

Answer 1

Answer :-

According to the Question

It is given that

• Height of object ,ho = 4cm
• Object distance ,u = -18cm
• focal length ,f = -12 cm
• Type of mirror = Concave mirror.

we have to calculate the position , nature and height of the image . Firstly we calculate the image distance of the object .By using mirror formula

• 1/v + 1/u = 1/f

where,

v denote image distance

u denote object distance

f denote focal length

Substitute the value we

➻ 1/v + 1/(-18) = 1/-12

➻ 1/v - 1/18 = -1/12

➻ 1/v = -1/12 + 1/18

➻ 1/v = -3+2/36

➻ 1/v = -1/36

➻ v = -36cm

• Hence, the image distance is 36 cm.

Now, calculating the height of image .

• hi/ho = -v/u

substitute the value we get

➻ hi/4 = -(-36)/18

➻ hi/4 = 2

➻ hi = 8

➻ hi = 8 cm

• Hence, the height of image is 8 cm
• Nature of Image :- Real and inverted .

Answer 2

Answer:

Given :-

• An object 4 cm height is placed at a distance of 18 cm from concave mirror having focal length 12 cm.

To Find :-

• What is the position, nature and height of the image.

Formula Used :-

$\clubsuit$ Mirror Formula :

$\mapsto \sf\boxed{\bold{\pink{\dfrac{1}{f} =\: \dfrac{1}{u} + \dfrac{1}{v}}}}$

where,

• f = Focal Length
• u = Object Distance
• v = Image Distance

$\clubsuit$ Magnification Formula :

$\mapsto \sf\boxed{\bold{\pink{\dfrac{h_i}{h_o} =\: \dfrac{- v}{u}}}}$

Solution :-

First, we have to find the image distance :

Given :

• Object Distance = - 18 cm
• Focal Length = - 12 cm

According to the question by using the formula we get,

$\implies \sf \dfrac{1}{- 12} =\: \dfrac{1}{- 18} + \dfrac{1}{v}$

$\implies \sf - \dfrac{1}{12} - \bigg(\dfrac{1}{- 18}\bigg) =\: \dfrac{1}{v}$

$\implies \sf - \dfrac{1}{12} + \dfrac{1}{18} =\: \dfrac{1}{v}$

$\implies \sf \dfrac{- 3 + 2}{36} =\: \dfrac{1}{v}$

$\implies \sf \dfrac{- 1}{36} =\: \dfrac{1}{v}$

By doing cross multiplication we get,

$\implies \sf - v =\: 36(1)$

$\implies \sf - v =\: 36$

$\implies \sf \bold{\purple{v =\: - 36\: cm}}$

Now, we have to find the height of image :

Given :

• Height of Object = 4 cm
• Image Distance = - 36
• Object Distance = 18

According to the question by using the formula we get,

$\longrightarrow \sf \dfrac{h_i}{4} =\: \dfrac{- (- 36)}{18}$

$\longrightarrow \sf \dfrac{h_i}{4} =\: \dfrac{36}{18}$

By doing cross multiplication we get,

$\longrightarrow \sf h_i \times 18 =\: 4 \times 36$

$\longrightarrow \sf 18h_i =\: 144$

$\longrightarrow \sf h_i =\: \dfrac{\cancel{144}}{\cancel{18}}$

$\longrightarrow \sf h_i =\: \dfrac{8}{1}$

$\longrightarrow \sf\bold{\red{h_i =\: 8\: cm}}$

${\small{\bold{\underline{\therefore\: The\: height\: of\: the\: image\: is\: 8\: cm\: .}}}}$

Hence, the nature of the image is real and inverted.