Question

Three lenses \(L_1\), \(L_2\) and \(L_3\), each of focal length \(40\) cm, are placed coaxially. The distance between \(L_1\) and \(L_2\) and between \(L_2\) and \(L_3\) are \(120\) cm and \(20\) cm respectively. An object is kept at a distance of \(80\) cm to the left of lens \(L_1\). Find the distance of the final image formed from the object.

Hint:

If an object is placed at \(2f\) of a convex lens, the image is also formed at \(2f\). If an object is placed at the principal focus of a convex lens, the emerging rays become parallel and the image is formed at infinity.

Answer 1

Concept: For each lens, the lens formula is \[ \frac{1}{f} = \frac{1}{v} -\frac{1}{u}. \] The image formed by one lens acts as the object for the next lens. Therefore, the problem is solved step-by-step by finding successive image positions.

Step 1: Image formed by lens \(L_1\). Given, \[ f_1=+40\text{ cm} \] \[ u_1=-80\text{ cm} \] Using lens formula, \[ \frac{1}{40} = \frac{1}{v_1} -\frac{1}{(-80)}. \] \[ \frac{1}{40} = \frac{1}{v_1} +\frac{1}{80}. \] \[ \frac{1}{v_1} = \frac{1}{40} -\frac{1}{80}. \] \[ \frac{1}{v_1} = \frac{1}{80}. \] \[ \boxed{v_1=80\text{ cm}} \] Thus the first image is formed \(80\) cm to the right of \(L_1\).

Step 2: Locate this image relative to lens \(L_2\). Distance between \(L_1\) and \(L_2\) \[ =120\text{ cm}. \] The image formed by \(L_1\) lies \[ 120-80=40\text{ cm} \] to the left of \(L_2\). Hence for lens \(L_2\), \[ u_2=-40\text{ cm}. \] \[ f_2=+40\text{ cm}. \] Applying lens formula, \[ \frac{1}{40} = \frac{1}{v_2} -\frac{1}{(-40)}. \] \[ \frac{1}{40} = \frac{1}{v_2} +\frac{1}{40}. \] Therefore, \[ \frac{1}{v_2}=0. \] \[ \boxed{v_2=\infty} \] Thus rays emerging from \(L_2\) become parallel.

Step 3: Formation of image by lens \(L_3\). Parallel rays fall on lens \(L_3\). For a convex lens, parallel rays converge at its principal focus. Therefore, \[ v_3=f_3. \] Since \[ f_3=40\text{ cm}, \] \[ \boxed{v_3=40\text{ cm}} \] Thus the final image is formed \(40\) cm to the right of \(L_3\).

Step 4: Determine the position of the final image relative to the object. Distance from \(L_1\) to \(L_3\) \[ =120+20 \] \[ =140\text{ cm}. \] Final image is \(40\) cm beyond \(L_3\). Hence distance of final image from \(L_1\) \[ =140+40 \] \[ =180\text{ cm}. \] The object is \(80\) cm to the left of \(L_1\). Therefore distance between object and final image is \[ 80+180. \] \[ \boxed{260\text{ cm}} \] Final Answer: \[ \boxed{ \text{Distance between object and final image} = 260\text{ cm} } \]