Question

An object and a screen are fixed 80cm apart.In one position for a convex lens a real image is formed on the screen wih a magnification of 2/3.Find the focal length of the lens.

Answer 1

Distance between object and image (image forms at screen) = 80 cm

Given that magnification = -2/3

For lenses,

v/u = -2/3

→ v = -2u/3

Distance between object and image, as states earlier, = 80 cm

→ v - u = 80

→ -2u/3 - u = 80

→ -5u/3 = 80

→ u = -48 cm

So, v = 32 cm

Lens formula gives

1/f = 1/v - 1/u

→ 1/f = 1/(32) - (1/-48)

→ $\sf\frac{1}{f} = \frac{1}{8}(\frac{1}{4} + \frac{1}{6}$

→ $\sf\frac{1}{f} = \frac{1}{8}(\frac{3}{12} + \frac{2}{12}$

$\sf\frac{1}{f} = \frac{1}{8}(\frac{5}{12})$

1/f = 5/96

f = 96/5 cm

→ f = 19.2 cm

Answer 2

GIVEN

An object and a screen are fixed 80 cm apart, i.e. the image is formed 80 cm apart from the object.

Magnification, m = 2/3

TO FIND

The focal length(f) of the convex lens.

SOLUTION

We know,

$\sf{M=\dfrac{v}{u} }$

Putting the values :-

$\sf{\dfrac{-2}{3} =\dfrac{v}{u} }\\\\\sf{\dashrightarrow -2u=3v }\\\\\sf{\dashrightarrow v=\dfrac{-2u}{3} }\:\:\:\: \cdots (i)$

We also know :-

$\displaystyle{\sf{\frac{1}{v}-\frac{1}{u}=\frac{1}{f} }}$

Putting the values :-

$\displaystyle{\sf{\dashrightarrow \frac{1}{\frac{-2u}{3} }-\frac{1}{u}=\frac{1}{f} }}\:\:\:\: \cdots \sf{[from\ (i)]}\\\\\displaystyle{\sf{\dashrightarrow \frac{-3}{2u}-\frac{1}{u}=\frac{1}{f} }} \\\\\displaystyle{\sf{\dashrightarrow \frac{-3-2}{2u}=\frac{1}{f} }} \\\\\displaystyle{\sf{\dashrightarrow \frac{-5}{2u}=\frac{1}{f} }} \\\\\displaystyle{\sf{\dashrightarrow f=\frac{-2u}{5} }}\:\:\:\: \cdots \sf{(ii)}$

A/q,

v - u = 80

$\sf{\implies \dfrac{-2u}{3} -u=80}\\\\\displaystyle{\sf{\implies \frac{-2u-3u}{3} =80}} \\\\\displaystyle{\sf{\implies -5u=240 }} \\\\\displaystyle{\sf{\implies u=-48\ cm }}$

Putting the value of "u" in eq.(ii) :-

$\displaystyle{\sf{ f=\frac{-2u}{5} }} \\\\\displaystyle{\sf{\dashrightarrow f=\frac{-2\times-48}{5} }} \\\\\displaystyle{\sf{\dashrightarrow f=\frac{96}{5} }} \\\\\boxed{\displaystyle{\sf{\dashrightarrow f= 19.2\ cm}} }$

Therefore, the focal length of the convex lens is 19.2 cm.