Question

an arrow 2,5 cm high is placed at distance of 25 cm from a diverging mirror of focal length of 20 cm . find the nature,position and size of the image formed.

Answer 1

Given :

➳ Height of arrow = 2.5cm

➳ Distance of object = 25cm

➳ Focal length = 20cm

➳ Type of mirror : convex

To Find :

◕ Nature of image

◕ Position of image

◕ Size of image

Solution :

➛ X-coordinate of centre of curvature and focus of concave mirror are negative and those for convex mirror are positive. In case of mirrors since light rays reflect back in X-direction, therefore -ve sign of v indicates real image and +ve sign of v indicates virtual image.

✴ Position of image :

$\longrightarrow\tt\:\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}\\ \\ \longrightarrow\tt\:\dfrac{1}{(-25)}+\dfrac{1}{v}=\dfrac{1}{20}\\ \\ \longrightarrow\tt\:\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{25}\\ \\ \longrightarrow\tt\:\dfrac{1}{v}=\dfrac{5+4}{100}\\ \\ \longrightarrow\tt\:\dfrac{1}{v}=\dfrac{9}{100}\\ \\ \longrightarrow\underline{\boxed{\bf{v=11.11cm}}}$

✴ Size of image :

$\longrightarrow\tt\:m=\dfrac{-v}{u}=\dfrac{h'}{h}\\ \\ \longrightarrow\tt\:\dfrac{-11.11}{-25}=\dfrac{h'}{2.5}\\ \\ \longrightarrow\tt\:h'=0.44\times 2.5\\ \\ \longrightarrow\underline{\boxed{\bf{h'=1.1cm}}}$

✴ Nature of image :

• virtual
• erect
• small

Answer 2

$\mathfrak{\huge{\underline{\underline{\red{Correct\:Question:-}}}}}$

✴ An arrow 2.5 cm high is placed at distance of 25 cm from a diverging mirror of focal length of 20 cm . Find the nature,position and size of the image formed.

$\mathcal{\huge{\fbox{\green{AnSwEr:-}}}}$

• The nature of image is virtual & erect .

• The image position will be formed at 11.11 cm behind the mirror .

• The size of the image formed is 1.11 cm .

$\mathcal{\huge{\fbox{\purple{Solution:-}}}}$

Given data :-

H1 = +2.5cm

u = -25cm

f = +20cm

To Find :-

• The nature of image.

• The position of image.

• The size of the image.

Calculation :-

✏ According to the mirror formula,

✺ 1/f = 1/v + 1/u

➠ 1/20 = 1/v -1/25

➠ 1/v = 1/20 + 1/25

➠ 1/v = 9/100

➠ v = 100/9

➠ v = +11.11 cm

☑ So, the position is +11.11 CM.

Using formulae of magnification ,

✯ m = h2/h1 = -v/u

➠ h2/2.5 = -11.11/-25

➠ h2 = -11.11/-25 × 2.5

➠ h2 = +1.11 cm

☑ So, the size is + 1.11 cm.

☆ A virtual and erect image of an object can be seen by diverging lens and mirror . The convex mirror and a concave lens, which are both diverging in nature can perform the task. So the image is virtual & erect.

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$\mathcal{\large{\fbox{\pink{Important\:Note:-}}}}$

an arrow 2,5 cm high is placed at distance of 25 cm from a diverging mirror of focal length of 20 cm . find the nature,position and size of the image formed.

➷➷➷➷➷➷➷➷

Avoid grammatical errors in question it confuses answerer & may give you wrong information.

Thank you.

Regards,

TheAdvancedHacker

❂ ComputerMaster

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