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 :

Given -

• Converging lens
• Object Distance = - 10 cm
• Focal length = + 15 cm
• Height of the object = + 2 cm

To Find -

• Ray Diagram.
• Three characteristics of the Image formed.

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

$\large{\red{ \bf{ \underline {\underline{Answer}}}}}$

$\purple{ \sf{\mapsto Height \: of \: the \: image = 6 \: cm}}$

$\mathrm{\green{ \underline{Given}}} \\ \\ \sf{ \rightsquigarrow Object \: distance \: (u)= - 10 \: cm} \\ \\ \sf{ \rightsquigarrow Focal \: length \: (f) = + 15 \: cm} \\ \\ \sf{ \rightsquigarrow Height\: of \:the\: object = + 2\: cm} \\ \\ \mathrm {\blue{ \underline{ To \: Find}}} \\ \\ \sf{ \rightsquigarrow Height \: of \: the \: image\:=\:?}$

⠀⠀

❏ According to given question

⠀⠀

$\bf{ \underline{From \: lens \: fromula}} \\ \\ \sf{ : \implies \frac{1}{f} = \frac{1}{v} - \frac{1}{u} } \\ \\ \sf{: \implies \frac{1}{v} = \frac{1}{f} + \frac{1}{u} } \\ \\ \sf{: \implies \frac{1}{v} = \frac{1}{15} - \frac{1}{10} } \\ \\ \sf{: \implies \frac{1}{v} = \frac{10 - 15}{150} } \\ \\ \sf{: \implies \frac{1}{v} = \frac{ \cancel{ - 5}}{ \cancel{150} } = \frac{ - 1}{30} } \\ \\ \sf{: \implies \frac{1}{v} = \frac{ - 1}{30} } \\ \\ \sf{ : \implies v = - 30 \: cm} \\ \\ \sf{ Magnification \: (m) = \frac{v}{u} = \frac{h_i}{h_o} } \\ \\ \sf{: \implies h_i = h_o \times \frac{v}{u} } \\ \\ \sf{ : \implies h_o = 2 \: cm} \\ \\ \sf{: \implies h_i = 2 \times \frac{ \cancel{-} \cancel{30 }}{ \cancel{ - } \cancel{10}}} \\ \\ \sf{: \implies h_i = 2 \times 3} \\ \\ \sf{ \purple{ : \implies} \purple{ \underline{ \boxed{ \sf{h_i = 6 \: cm}}}}}$

❍ Image forms on the same side as that of object.

❍ Image is virtual and erect.

❍ image is 3 times larger than height of object.