Question

The image formed by a spherical mirror is real inverted and is magnification -1 if the image is a distance of 30 from the mirror. where is the object placed. find the position of the image if the object is moved 20 cm toward the mirror. what is the nature

Answer 1

Solution:

• Magnification (m) = - 1
• Distance of the image from the mirror (v) = - 30
• Distance of the object from the mirror (u) = ? (to be calculated)

Applying,

m = - v/u

➾ - 1 = - (- 30)/u

➾ u = - 30/1

➾ u = - 30

Therefore, Distance of the object from the mirror (u) = - 30 cm

If object is placed at a distance of 30 cm from the mirror,

➾ 1/f = 1/v - 1/u

➾ 1/f = - 1/30 - 1/30

➾ 1/f = - 2/30

➾ f = - 30/2

➾ f = - 15 cm

Now,

If the object is moved 20 cm towards the mirror (u = 30 - 20 = 10 cm ), the position of the image or image distance becomes,

➾ 1/f = 1/v + 1/u

➾ - 1/15 = 1/v + ( - 1/10 )

➾ - 1/15 + 1/10 = 1/v

➾ 1/30 = 1/v

➾ v = 30 cm

Therefore, the image is formed 30 cm behind the mirror.

Now,

➾ m = - v/u

➾ m = - 30/-10

➾ m = 3

Magnification is positive, therefore the image is virtual, erect and enlarged.

_______________________

Answer 2

Answer:

Given :-

• The image formed by a spherical mirror is real inverted and it's magnification is - 1.
• The image is a distance of 30 cm from the mirror.
• The object is moved 20 cm towards the mirror.

To Find :-

• What is the position of the image.
• What is the nature of the image.

Formula Used :-

$\clubsuit$ Magnification Formula :

$\longrightarrow \sf\boxed{\bold{\pink{Magnification =\: \dfrac{- v}{u}}}}$

where,

• v = Image Distance
• u = Object Distance

$\clubsuit$ Mirror Formula :

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

where,

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

Solution :-

First, we have to find the object distance :-

Given :

• Magnification (m) = - 1
• Image Distance (v) = - 30 cm

According to the question by using the formula we get,

$\leadsto \sf - 1 =\: \dfrac{- (- 30)}{u}$

$\leadsto \sf - 1 =\: \dfrac{30}{u}$

By doing cross multiplication we get,

$\leadsto \sf - 1 \times u =\: 30$

$\leadsto \sf - u =\: 30$

$\leadsto \sf\bold{\purple{u =\: - 30}}$

Now, we have to find the focal length :-

Given :

• Object Distance = - 30 cm
• Image Distance = - 30 cm

According to the question by using the formula we get,

$\implies \sf \dfrac{1}{f} =\: \bigg(- \dfrac{1}{30}\bigg) + \bigg(- \dfrac{1}{30}\bigg)$

$\implies \sf \dfrac{1}{f} =\: - \dfrac{1}{30} - \dfrac{1}{30}$

$\implies \sf \dfrac{1}{f} =\: \dfrac{- 1 - 1}{30}$

$\implies \sf \dfrac{1}{f} =\: \dfrac{- \cancel{2}}{\cancel{30}}$

$\implies \sf \dfrac{1}{f} =\: \dfrac{- 1}{15}$

By doing cross multiplication we get,

$\implies \sf - 1 \times f =\: 15$

$\implies \sf - f =\: 15$

$\implies \sf\bold{\green{f =\: - 15\: cm}}$

Now, we have to find the new image distance :-

$\bigstar$ When object is moved 20 cm towards the mirror.

$\leadsto \sf u =\: 30 - 20$

$\implies \bf u =\: 10\: cm$

Given :

• Focal Length (f) = - 15 cm
• Object Distance (u) = - 10 cm

According to the question by using the formula we get,

$\dashrightarrow \sf \dfrac{1}{f} =\: \dfrac{1}{v} + \dfrac{1}{u}$

$\dashrightarrow \bf \dfrac{1}{f} - \dfrac{1}{u} =\: \dfrac{1}{v}$

$\dashrightarrow \sf \bigg\{- \dfrac{1}{15}\bigg\} - \bigg\{- \dfrac{1}{10}\bigg\} =\: \dfrac{1}{v}$

$\dashrightarrow \sf - \dfrac{1}{15} + \dfrac{1}{10} =\: \dfrac{1}{v}$

$\dashrightarrow \sf \dfrac{- 2 + 3}{30} =\: \dfrac{1}{v}$

$\dashrightarrow \sf \dfrac{1}{30} =\: \dfrac{1}{v}$

By doing cross multiplication we get,

$\dashrightarrow \sf 1 \times v =\: 1 \times 30$

$\dashrightarrow \sf\bold{\red{v =\: 30\: cm}}$

$\therefore$ The position of the image is 30 cm behind the mirror.

Again,

Given :

• Image Distance (v) = 30 cm
• Object Distance (u) = - 10 cm

According to the question by using the formula we get,

$\implies \bf m =\: \dfrac{- v}{u}$

$\implies \sf m =\: \dfrac{\cancel{-} 30}{\cancel{-} 10}$

$\implies \sf m =\: \dfrac{3\cancel{0}}{1\cancel{0}}$

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

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

Hence, the magnification is positive the image is erect and virtual.

$\therefore$ The nature of the image is virtual, erect and enlarged.

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