The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 4 m away by means of a large convex lens. The maximum possible focal length of the lens required for this purpose will be:

Question

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 4 m away by means of a large convex lens. The maximum possible focal length of the lens required for this purpose will be: (A) 0.5 m (B) 1.0 m (C) 1.5 m (D) 2.0 m

Tardigrade Admin · Accepted Answer

Let a large convex lens is placed between two walls at a distance x from wall on which an electric bulb is fixed. Using lens formula,

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

Putting

u = x

and

v = 4 - x

∴

\frac{1}{f} = \frac{1}{4 - x} - \frac{1}{- x}

Or

\frac{1}{f} = \frac{x + 4 - x}{( 4 - x ) ( x )}

Or

\frac{1}{f} = \frac{4}{( 4 - x ) ( x )}

Or

f = \frac{( 4 - x ) ( x )}{4}

.. (i)
Now magnification,

m = \frac{v}{u} = \frac{4 - x}{x}

or

1 = \frac{4 - x}{x}

or

x = 4 - x

or

2 x = 4

or

x = 2 m

Hence, from equation (i),

f = \frac{( 4 - 2 ) ( 2 )}{4} = \frac{2 \times 2}{4} = 1 m

Q. The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 4m away by means of a large convex lens. The maximum possible focal length of the lens required for this purpose will be:

0.5 m

1.0 m

1.5 m

2.0 m

Q. The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall $4 m$ away by means of a large convex lens. The maximum possible focal length of the lens required for this purpose will be: