Question

An object of height 5cm is placed perpendicular to the principal axis of a concave lens of

focal length 10cm. if the distance of the object from the optical centre of the lens is 20cm;

determine the position, nature and size of the image.

Answer 1

According to the Question

It is given that

• Object of height ,ho = 5cm
• Focal length ,f = -10cm.
• Object distance ,u = -20cm

we have to calculate the the position, nature and size of the image. Firstly we calculate the image distance .

By using Lens Formula.

• 1/f = 1/v - 1/u

by putting the value we get

↠ 1/v = 1/f + 1/u

↠ 1/v = 1/-10 + 1/-20

↠ 1/v = -1/10 - 1/20

↠ 1/v = -1 +(-2)/20

↠ 1/v = -1-2/20

↠ 1/v = -3/20

↠ v = -20/3

↠ v = -6.67 cm

• Hence, the image distance is 6.67cm from the object and the image is formed in front of mirror.

Now,

Calculating magnification

• m = hi/ho = v/u

by putting the value we get

↠ hi/5 = -6.67/(-20)

↠ hi = -6.67×5/(-20)

↠ hi = 33.35/20

↠ hi = 1.66 cm

• Hence, the height of image is 1.66 cm .
• The image formed is smaller than the object and erect .

Answer 2

$\large\underline{\underline{\maltese{\color{darkblue}{\pmb{\sf{\: Given :-}}}}}}$

➬ An object of height (h1)

• = 5 cm

➬ Focal length of concave lense = -10 cm

➬ Object Distance (u)

• = -20 cm

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\large\underline{\underline{\maltese{\pink{\pmb{\sf{\: To \: Find :-}}}}}}$

➬ Image Distance (v) =?

➬Nature of Image = ?

➬ Image Height (h2)= ?

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\large\underline{\underline{\maltese{\orange{\pmb{\sf{\: Solution :-}}}}}}$

❒ Formula Used :

➢According to lens Formula

$\large{\blue{\bigstar}}{\underline{\boxed{\red{\sf{ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} }}}}}$

➳ Place the given value in this formula

$\large{\blue{\bigstar}}{\underline{\boxed{\pink{\sf{ \frac{1}{v} - \frac{1}{ - 20} = \frac{1}{ - 10} }}}}}$

$\large{\blue{\bigstar}}{\underline{\boxed{\blue{\sf{ \frac{1}{v} + \frac{1}{ 20} = \frac{ - 1}{ 10} }}}}}$

$\large{\blue{\bigstar}}{\underline{\boxed{\green{\sf{ \frac{1}{v} = \frac{ - 1}{10} - \frac{ 1}{ 20} = \frac{ - 2 - 1}{20} = \frac{ - 3}{20} }}}}}$

$\large{\blue{\bigstar}}{\underline{\boxed{\purple{\sf{v = \frac{ - 20}{3} = - 6.67 }}}}}$

$\large\qquad{\color{maroon}{:\longmapsto{\underline{\boxed{\sf{Image \: \: Distance = -6.67 \: cm}}}}}}{\pink{\bigstar}}$

$\qquad{━━━━━━━━━━━━━━━}$

❒ Note :

• The image to be formed on same side of object is virtual and negative sign (-) for image Distance (v) indicate that image is formed on left side of concave lens .
• The image is diminished because

${:\longmapsto{\sf{ |v| < |u| }}}$

❒ Now, Formula Used :

➢According to magnification Formula :

$\large\qquad{\color{green}{:\longmapsto{\underline{\boxed{\sf{Magnification }}}}}}{\pink{\bigstar}}$

$\large{\blue{\bigstar}}{\underline{\boxed{\red{\sf{m = \frac{v}{u} ,m = \frac{h1}{h2} }}}}}$

$\large{\blue{\bigstar}}{\underline{\boxed{\pink{\sf{ \implies\ \ \frac{ \frac{ - 20}{3} }{ - 20 } = \frac{h1}{5} }}}}}$

$\large{\blue{\bigstar}}{\underline{\boxed{\purple{\sf{h1 = \frac{5}{3} }}}}}$

$\large\qquad{\color{maroon}{:\longmapsto{\underline{\boxed{\sf{Image \: \: Size = +1.66 \: cm}}}}}}{\pink{\bigstar}}$

❒ Note :

• The postive sign (+) of Magnification h2 indicate that the image to be formed on same side of object is erect and virtual
• The image is diminished because

${:\longmapsto{\sf{ h2 < h1 }}}$

❒ Therefore :

❝ The image formed is 6.66 cm away from the lens on same side of object ❞

$\qquad{━━━━━━━━━━━━━━━}$

${\blue{\underline{▬▬▬▬▬}}{\pink{\underline{▬▬▬▬▬}}{\orange{\underline{▬▬▬▬▬}}{\purple{\underline{▬▬▬▬▬▬}}}}}}$