Question

An object is placed at the following distances from a concave mirror of focal length 15 cm, turn by turn
(a) 35 cm
(b) 30 cm
(c) 20 cm
(d) 10 cm
Which position of the object will produce
(i) a magnified real image?
(ii) a magnified virtual image?
(iii) a diminished real image?
(iv) an image of same size as the object​

Answer 1

$\huge\pink{\fbox{\tt{࿐αɴѕωєя࿐}}}$

$Given :$

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

Hope this above answer may help you

Answer 2

Answer:

Solution :

In this case the focal length of concave mirror is 10 cm. This means that the focus (F) of this concave mirror is at a distanace of 10 cm from the mirror and its centre of curvature ( C) is at a distance of 2×10=20cm form the mirror.

(i) A diminished real image is formed by a concave morror when the object is beyond C. Here C is at 20 cm. So, the diminished real image will be formed when the object is at a distance greater then 20 cm, which in this problem is 25cm. So, the position of object for a diminished real image is 25 cm.

(ii) A magnified real image is formed by a concave mirror when the object is between F and C. Here F is at 10 cm and C is at 20 cm. So, the magnified real image will be formed when the object is at a distance btween 10 cm and 20 cm, which in this problem is 15 cm. So, the position of object for a magnified real image is 15 cm.

(iii) A magnified virtual image is formed by a concave mirror when the object is within focus (F) at a distance less than focal length or less than 10 cm, which in this problem is 8 cm. So, the position of object for ma magnified virtual image is 8 cm.

(iv) An image of the same size as object is formed by a concave mirror when the object is at centre of curvature ( C). Here C is at 20 cm. So, the image of same as the object will be formed when the object is at 20 cm form the concave mirror. Thus, the position of object for an image of same size as the object is 20 cm.

Please note that we have given so many details in answering the above question just to make you understand clearly. There is, no need to give so many details in the examination. In the examination, the answer to the above question can just be written as :

(i) 25 cm (ii) 15 cm (iii) 8 cm (iv) 20 cm