Question

Answer the above attachment​

Answer 1

Answer.

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 :

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

✴ Size of image :

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

✴ Nature of image :

• ✴ Nature of image :virtual
• ✴ Nature of image :virtualerect

✴ Nature of image :virtualerect small

Answer 2

${\huge{\underline{\large{\mathbb{\red{ ᥫ᭡KABIR♡~}}}}}}$

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 :

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

Size of image :

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

Nature of image :

• Nature of image :virtual
• Nature of image :virtualerect

Nature of image :virtualerect small

Hope this helps you ❤❤