Solveeit Logo

Question

Question: The distance between two point sources of light is 24cm. Find out where you would place a converging...

The distance between two point sources of light is 24cm. Find out where you would place a converging lens of focal length 9 cm9{\text{ }}cm, so that the images of both the sources are formed at the same point.

Explanation

Solution

For lens the distance of the image formed from the lens is given by 1v  1u = 1f\dfrac{1}{v}{\text{ }} - {\text{ }}\dfrac{1}{u}{\text{ }} = {\text{ }}\dfrac{1}{f} where u, vu,{\text{ }}v and ff are a distance of the object, a distance of image, and the focal length respectively.

The focal length is the distance from the center of the lens to the principal foci or the focal points of the lens. For a converging lens, the best example is a convex lens. The focal length is positive and the interval at which a beam of collimated light will be focused on a single spot.

Complete step by step solution:

S1{S_1}we have to take the equation as
1v11x=19\dfrac{1}{{{v_1}}} - \dfrac{1}{{ - x}} = \dfrac{1}{9}
Then the equation becomes
1v1+1x=19\dfrac{1}{{{v_1}}} + \dfrac{1}{x} = \dfrac{1}{9}
Now we want the value of 1v1\dfrac{1}{{{v_1}}} so take it outside
1v1=191x.........(1)\dfrac{1}{{{v_1}}} = \dfrac{1}{9} - \dfrac{1}{x}.........(1)
Now forS2{S_2}we have to take the equation as
1v21(24x)=19\dfrac{1}{{{v_2}}} - \dfrac{1}{{ - (24 - x)}} = \dfrac{1}{9}
Then the equation becomes
1v2+1(24x)=19\dfrac{1}{{{v_2}}} + \dfrac{1}{{(24 - x)}} = \dfrac{1}{9}
Now we want the value of 1v2\dfrac{1}{{{v_2}}} so take it outside
1v2=191(24x)\dfrac{1}{{{v_2}}} = \dfrac{1}{9} - \dfrac{1}{{(24 - x)}}
Now Since, the sign convention for S1{S_1} and S2{S_2}is just opposite. Hence, we should take
v1=v2{v_1} = - {v_2}
Then the equation becomes
1v1=1v2\Rightarrow \dfrac{1}{{{v_1}}} = - \dfrac{1}{{{v_2}}}
Therefore now substitute the values so we get
191x=124x19\dfrac{1}{9} - \dfrac{1}{x} = \dfrac{1}{{24 - x}} - \dfrac{1}{9}
Solving this equation, we get x=6 cmx = 6{\text{ }}cm. Therefore, the lens should be kept at a distance of 6 cm6{\text{ }}cm from either of the objects.

Note: A converging lens is defined as a lens that converges rays of the light that are moving parallel to the principal axis. The fact that a double convex lens is thicker across its middle is an indicator that it will converge rays of light that move parallel to its principal axis.