Question

An object is placed 10cm from a converging lens of focal length 15cm. If the height of the object is 2cm, calculate height of the image. Draw the necessary ray diagram. Write three characteristics of the image formed.​

Answer 1

Answer:

Solution -

1/v - 1/u = 1/f

⇒ 1/v - 1/-10 = 1/15

⇒ 1/v + 1/10 = 1/15

⇒ 1/v = 1/15 - 1/10

⇒ 1/v = (10 - 15)/150

⇒ 1/v = -5/150

⇒ 1/v = -1/30

⇒ -v = 30

⇒ v = -30 cm

Image Distance = -30 cm.

M = h'/h = v/u

⇒ h'/2 = -30/-10

⇒ h'/2 = 3

⇒ h' = 6 cm

Height of the Image = 6 cm.

Characteristics of the Image :

Enlarged [Size]

Virtual and Erect [Nature]

On the same side of the lens as the object [Position]

Answer 2

Figure :-

$\setlength{\unitlength}{1.6mm}\begin{picture}(5,7)\qbezier(0,20)(7,8)(0,0)\qbezier(0,20)(-7,8)(0,0)\put(-25,8){\line(1,0){48}}\put(-22,8){\vector(0,1){11.5}}\put(10,3){\line(-2,1){32}}\put(16,6){\line(-3,1){38}}\put(0,8){\circle*{0.7}}\put(-1,5){\huge{o}}\put(-7.5,8){\vector(0,1){4}}\put(-11,8){\circle*{0.7}}\put(9.8,8){\circle*{0.7}}\put(-18,8){\circle*{0.7}}\put(-20,6){2F1}\put(-11,6){F1}\put(17,8){\circle*{0.7}}\put(17,6){2F2}\put(9,6){F2}\put(-7.5,12){\vector(1,0){2}}\put(-7.5,12){\line(1,0){5.2}}\end{picture}$

$\huge{\red{ \textbf{Solution\::-}}}$

An object of height of 2 cm is placed at 10 cm away from a converging lens, having a focal length of 15 cm. What we need to find is the height of the image (v).

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

➯ $\sf \dfrac{1}{15} = \dfrac{1}{v} - \dfrac{1}{-10}$

➯ $\sf \dfrac{1}{15} - \dfrac{1}{10} = \dfrac{1}{v}$

➯ $\sf \dfrac{10 -15}{150 } = \dfrac{1}{v}$

➯ $\sf \dfrac{-5}{150 } = \dfrac{1}{v}$

➯ $\sf \dfrac{-1}{30} = \dfrac{1}{v}$

➯ $\sf -30 \:cm = v$

Image distance = -30 cm.

➯ $\sf \dfrac{h1}{h0} = \dfrac{v}{u}$

➯ $\sf \dfrac{h1}{2} = \dfrac{-30}{-10}$

➯ $\sf h1 = \dfrac{-60}{-10}$

➯ $\sf h1= 6 \:cm$

Height of the image = 6 cm.

${\red{ \textbf{Characterstics\:of\:image\::-}}}$

• Enlarged.
• Virtual and erect.
• Image forms on same side as that of the object.