Question

a object is placed at 25 focal length 60 what image distance convex lens ​

Answer 1

$\huge\sf{Answer:}$

☆ Given:

⇏A object is placed at 25 focal length 60.

☆ Find:

⇏ What is the image distance.

☆ Know terms:

⇏ Object distance = (u)

⇏ Focal length = (f)

⇏ Image distance = (v)

☆ Using formula: Lens formula:

⇏ $\sf \dfrac{1}{v} - \dfrac{1}{u} + \dfrac{1}{f}$

☆ Calculations:

⇏ $\sf \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$

⇏ $\sf \dfrac{1}{v} = \dfrac{1}{60} + \dfrac{1}{25}$

⇏ $\sf \dfrac{1}{v} = [\dfrac{25 + (-60)}{1500}]$

⇏ $\sf \dfrac{25-60}{1500}= \dfrac{-35}{1500}$

⇏ $\sf \dfrac{-300}{7}$

⇏ $\sf \dfrac{-1500}{35}$

⇏ ${\sf{\boxed{\sf{-42.85 cm}}}}$

Therefore, -42.85 cm is the image distance.

Answer 2

Answer:

Given :

• Focal length = 60 cm
• Convex lens
• Object distance = -25 cm

To Find :

• Image distance ?

Solution :

Lens formula : 1/f = 1/v - 1/u

➸ 1/v = 1/f + 1/u

➸ 1/v = 1/60 + 1/-25

➸ 1/v = (-25 + 60)/1500

➸ 1/v = (60 - 25)/1500

➸ 1/v = 35/-1500

➸ v = -1500/35 = -300/7

➸ v = -42.85 cm

So, the image distance = -42.85 cm

Note : The ray diagram is in attachment.

EXTRA :

➸ The image formed will be virtual and erect. [Nature of the image]

➸ The image will be enlarged. [Size]

➸ The image will be formed on the same side of the lens as the object. [Position]