Physics, asked by bickyrajsoibam, 1 year ago

Practice Problems : 1
1. An object when placed in front of a convex lens forms a real image of 0.5 magnification. If the
distance of the image from the lens is 24 cm, calculate the focal length of the lens .

Answers

Answered by Anonymous

$<html><head><style>h1 {text-shadow: 0 0 11px #FF0000;}</style></head><body><h1> ♔ Vital Contributor </h1></body></html>$

$\mathcal{\huge{\underline{\underline{\red{Given:-}}}}}$

$\mathcal{\huge{\underline{\underline{\green{To\:Find:-}}}}}$

⬛ The focal length of the lens.

$\mathcal{\huge{\underline{\underline{\blue{Answer:-}}}}}$

$\mathcal{\huge{\underline{\underline{\pink{Solution:-}}}}}$

▶Magnification, m=−0.5 [ Image is real ]

▶v = 24 cm

▶ $m = \dfrac{v}{u}$

▶ $- 0.5 = \dfrac{24}{u}$

Using,

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

➡ $\dfrac{1}{24} - \dfrac{1}{( - 48)} = \dfrac{1}{f}$

➡ $\dfrac{1}{f} = \dfrac{1}{24} + \dfrac{1}{48}$

➡ $\dfrac{1}{f} = \dfrac{2 + 1}{48}$

➡ $\dfrac{1}{f} = \dfrac{3}{48}$

➡ $f = \cancel{ \dfrac{48}{3}}$

✒ $f \: = 16 \: cm$ ✔

________________________________

Answered by Anonymous

Answer:

$Answer$

Mark the upper budy brainliest dude.

Previous Question

Next Question