Question

18) The focal length of a convex lens 1850cm.
a obiect he kept at a distance of 75cm.
from the lens, then find the smage
distance and magnification produced by
the lens.​

Answer 1

Question :-

The focal length of a convex lens is 1850 cm .An object is kept at a distance of 75 cm from the lens

,then find the image distance and magnification produced by lens .

Answer :-

$\mapsto(1) \: \: \boxed{ \sf \: v = - 78.17 \: cm} \: \\ \\ \mapsto \: (2) \: \boxed{\sf m = 1.042 \: \: \: \: \: \: \: }\\ \:$

Explanation :-

According to the question ,

• → Object distance (u) = 75 cm

• → Focal length (f) = 75 cm

• → image distance (v) = ?

• → Magnification power (m) = ?

We know that ,

→ In convex lens we will take object distance is in negative direction because direction of incident ray is positive .

Solution :-

Apply the lens maker formula :

$\mapsto \sf \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \\ \to \sf \: \frac{1}{1850} = \frac{1}{v} - \frac{1}{ - 75} \\ \\ \to \sf \: \frac{1}{v} = \frac{1}{1850} - \frac{1}{75} \\ \\ \mapsto \sf \frac{1}{v} = \frac{ - 1775}{138750} \\ \\ \mapsto \sf \: v = - \frac{138750}{1775} \\ \\ \mapsto \boxed{ \sf \: v = - 78.17 \: cm}$

hence image is formed left side of lens .

Now ,we know that

→ Magnification power of convex lens :

$\to \sf \: m \: = \frac{v}{u} \\ \\ \to \sf m = \sf \: \frac{ - 78.17}{-75} \\ \\ \to \boxed{\sf \: m = 1.042}\\ \\ \sf \: \:$

→ Because magnification is positive hence image is inverted .

→ m > 1 ,so image is larger then the size of object .

Answer 2

Solution:

Given:

• Focal length of convex lens = 1850 cm
• Object Distance = 75 cm

To Find:

• Image distance
• Magnification

By using lens formula, we get

$\implies \sf \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\ \\ \\ \implies \sf \dfrac{1}{1850}=\dfrac{1}{v}-\dfrac{1}{(-75)}\\ \\ \\ \implies \sf \dfrac{1}{1850}-\dfrac{1}{75}=\dfrac{1}{v} \\ \\ \\ \implies \sf \dfrac{3-71}{5550} = \dfrac{1}{v}\\ \\ \\ \implies \sf \dfrac{-71}{5550}=\dfrac{1}{v}\\ \\ \\ \implies -71v=5550\\ \\ \\ \implies \sf v=-\dfrac{5550}{71}\\ \\ \\ \implies {\boxed{\red{\sf v=-78.16\;cm}}}$

Hence, Image distance = -78.16 cm.

$\rule{100}{1}$

Now we will calculate magnification,

$\implies \sf Magnification=\dfrac{v}{u}\\ \\ \\ \implies \sf m = \dfrac{-78.16}{-75}\\ \\ \\ \implies \sf m = 1.04 \\ \\ \\ \implies{\red{\boxed{\sf m=1.04}}}$

Hence, Magnification = 1.04.