Question

An object is placed at the following distances from a concave mirror of focal length 10cm (a) 8cm (b) 15 cm (c) 20 cm (d) 25 cm

Which position of object will produce
(i) A diminished real image ?
(ii) A magnified real image ?
(iii) A magnified virtual image ?
(iv) An image of same size as
the object ?

Answer 1

Answer:

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Answer 2

$\huge\pink{\fbox{\tt{࿐αɴѕωєя࿐}}}$

Given :

An object is placed at following distances from a concave mirror of focal length 15 cm.

To Find :-

➦ Which position of the object will produce :

(i) Virtual image.

(ii) Diminished and real image.

(iii) Magnified and real image.

(iv) Image of same size.

Formula Used :-

$\longmapsto \sf\boxed{\bold{\pink{m =\: \dfrac{f}{f - u}}}}$

where,

m = Magnification

m = Magnificationf = Focal length

m = Magnificationf = Focal lengthu = Object distance

m = Magnificationf = Focal lengthu = Object distanceSolution :-

(a) Object distance = 10 cm

Given :

Object distance = 10 cm

Focal length = 15 cm

According to the question by using the formula we get:

$\implies \sf m =\: \dfrac{15}{15 - 10}$

$\implies \sf m =\: \dfrac{\cancel{15}}{\cancel{5}}$

$\implies \sf m =\: \dfrac{3}{1}$

$\implies\sf\bold{\red{m =\: 3}}$

$\longrightarrow \sf\bold{\purple{Magnification =\: 3}}$

${\small{\bold{\green{\underline{\therefore It\: is\: Virtual\: Image\: .}}}}}$

$\rule{300}{2}$

(b) Object distance = 20 cm

Given :

Object distance = 20 cm

Focal length = 15 cm

According to the question by using the formula we get :

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

$\implies \sf m =\: \dfrac{\cancel{15}}{- \cancel{5}}$

$\implies \sf m =\: \dfrac{3}{- 1}$

$\implies \sf\bold{\red{m =\: - 3}}$

$\longrightarrow \sf\bold{\purple{Magnification =\: - 3}}$

${\small{\bold{\green{\underline{\therefore It\: is\: Magnified\: and\: Real\: image\: .}}}}}$

$\rule{300}{2}$

(c) Object distance = 30 cm

Given :

Object distance = 30 cm

Object distance = 30 cmFocal length = 15 cm

According to the question by using the formula we get :

$\implies \sf m =\: \dfrac{15}{15 - 30}$

$\implies \sf m =\: \dfrac{\cancel{15}}{-\cancel{15}}$

$\implies \sf m =\: \dfrac{1}{- 1}$

$\implies \sf\bold{\red{m =\: - 1}}$

$\longrightarrow \sf\bold{\purple{Magnification =\: - 1}}$

${\small{\bold{\green{\underline{\therefore It\: is\: Image\: Of\: Same\: Size\: .}}}}}$

$\rule{300}{2}$

d) Object distance = 40 cm

d) Object distance = 40 cmGiven :

d) Object distance = 40 cmGiven :Object distance = 40 cm

Focal length = 15 cm

According to the question by using the formula we get :

$\implies \sf m =\: \dfrac{15}{15 - 40}$

$\implies \sf m =\: \dfrac{\cancel{15}}{- \cancel{25}}$

$\implies \sf m =\: \dfrac{3}{- 5}$

$\implies \sf\bold{\red{m =\: - 0.6}}$

$\longrightarrow \sf\bold{\purple{Magnification =\: - 0.6}}$

${\small{\bold{\green{\underline{\therefore It\: is\: Diminished\: and\: real\: image\: .}}}}}$

$\rule{300}{2}$

The position of the object will produce :

(i) Virtual image $\dashrightarrow$ (a) 10 cm

(ii) Diminished and real image $\dashrightarrow$ (d) 40 cm

(iii) Magnified and real image $\dashrightarrow$ (b) 20 cm

(iv) Image of same size $\dashrightarrow$ (c) 30 cm

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