Question

Answer the question
with steps​

Answer 1

Given :

• Focal length (f) = -20cm
• Distance of the image (v) = -15cm
• height of the object $\sf (h_o)$ = 5cm

To Find :

• Distance of the object (u)
• height of the image $\sf (h_i)$

Formulas Applied :

Case 1 :-

$\Rightarrow \sf \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

$\underline{\boxed{\sf \dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}}}$

Case 2 :-

$\sf magnification = \dfrac{height_{(image)}}{height_{(object)}}$ _____(1)

$\sf magnification = \dfrac{distance _{(image)}}{distance _{(object)}}$ _____(2)

On equating the equations (1) and (2)

$\sf \dfrac{height_{(image)}}{height_{(object)}}=\dfrac{distance _{(image)}}{distance _{(object)}}$

$\underline{\boxed{\sf \dfrac{h_i}{h_o}=\dfrac{v}{u}}}$

Solution :

1. Distance of the Object

$\longrightarrow \sf \dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}\\\\\longrightarrow \sf \dfrac{1}{u} = \Big(\dfrac{1}{-15}\Big)-\Big(\dfrac{1}{-20}\Big)\\\\\longrightarrow \sf \dfrac{1}{u}=\dfrac{-1}{15}+\dfrac{1}{20}\\\\\longrightarrow \sf \dfrac{1}{u}=\dfrac{-20+15}{300}\\\\\longrightarrow \sf \dfrac{1}{u}=\dfrac{-5}{300}\\\\\longrightarrow \sf \dfrac{1}{u}=\dfrac{-1}{60}\\\\\boxed{\sf u = -60\:cm}$

2. height of the image

$\longrightarrow \sf \dfrac{h_i}{h_o}=\dfrac{v}{u}\\\\\longrightarrow \sf \Big(\dfrac{h_i}{5} \Big)=\Big(\dfrac{-15}{-60}\Big)\\\\\longrightarrow \sf h_i = \dfrac{15 \times {\cancel{5}}^1}{{\cancel{60}}_{12}}\\\\\longrightarrow \sf h_i = \dfrac{15}{12} \Rightarrow \dfrac{5}{4}\\\\\boxed{\sf h_i = 1.25 \: cm}$

Required Answer :

Distance of the object is $\underline{\sf - 60\:cm}$

Height of the image is $\underline{\sf 1.25 \:cm}$