As shown in the figure, a combination of a thin plano-concave lens and a thin plano-convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is 30 cm and refraction index of the material for both the lenses is 1.75. Both the lenses are placed at distance of 40 cm from each other. Due to the combination, the image of the object is formed at distance $$x =$$ ______ cm, from concave lens.

Lens maker formula used:

$$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

For the plano-concave lens,

$$\frac{1}{f_1}=(1.75-1)\left(-\frac{1}{30}\right)$$

$$\frac{1}{f_1}=0.75\left(-\frac{1}{30}\right)$$

$$f_1=-40\ \text{cm}$$

For the plano-convex lens,

$$\frac{1}{f_2}=(1.75-1)\left(\frac{1}{30}\right)$$

$$\frac{1}{f_2}=0.75\left(\frac{1}{30}\right)$$

$$f_2=40\ \text{cm}$$

The object is at infinity, so after refraction from the concave lens ($$L_1$$), the image is formed at its focal point.

Since $$f_1=-40$$ cm, the image formed by $$L_1$$ is virtual and $$40$$ cm to the left of $$L_1$$.

Distance between lenses:

$$L_1L_2 = 40\ \text{cm}$$

Therefore, object distance for $$L_2$$:

$$u_2 = 40+40=80\ \text{cm}$$

Since,

$$u_2=2f_2$$

Now we use the standard lens property:

When an object is placed at $$2f$$ of a convex lens, the image is formed at $$2f$$ on the other side.

So,

$$v_2=2f_2=80\ \text{cm}$$

You can also verify using lens formula:

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

Putting,

$$f=40,\qquad u=-80$$

$$\frac{1}{40}=\frac{1}{v}+\frac{1}{80}$$

$$\frac{1}{v}=\frac{1}{40}-\frac{1}{80} = \frac{1}{80}$$

$$v=80\ \text{cm}$$

So image is formed at $$80\ \text{cm}$$ on the right side of the convex lens.

Hence total distance from concave lens:

x = 40 + 80

$$\boxed{x=120\ \text{cm}}$$