An object $$2.4$$ m in front of a lens forms a sharp image on a film $$12$$ cm behind the lens. A glass plate $$1$$ cm thick, of refractive index $$1.50$$ is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film?

Solution & Explanation

Let us analyze the initial configuration before the glass slab is introduced. The object distance ($$u_1$$) and image distance ($$v_1$$) are given as:

$$u_1 = -2.4 \,\, \text{m} = -240 \,\, \text{cm}$$

$$v_1 = +12 \,\, \text{cm}$$

Using the standard thin lens formula, we find the focal length ($$f$$) of the lens:

$$\frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1}$$

$$\frac{1}{f} = \frac{1}{12} - \frac{1}{-240} = \frac{1}{12} + \frac{1}{240}$$

$$\frac{1}{f} = \frac{20 + 1}{240} = \frac{21}{240} \implies f = \frac{240}{21} \,\, \text{cm}$$

When a glass plate of thickness $$t = 1 \,\, \text{cm}$$ and refractive index $$\mu = 1.50$$ is interposed between the lens and the film, it shifts the optical path of the rays traveling toward the image focus. The normal shift ($$\Delta x$$) caused by this parallel-faced slab is:

$$\Delta x = t \left(1 - \frac{1}{\mu}\right)$$

$$\Delta x = 1 \cdot \left(1 - \frac{1}{1.5}\right) = 1 \cdot \left(1 - \frac{2}{3}\right) = \frac{1}{3} \,\, \text{cm}$$

Because the slab shifts the incoming converging rays forward by $$\frac{1}{3} \,\, \text{cm}$$, the lens must now form its primary image slightly closer to the lens so that the final shifted focus lands perfectly on the stationary film located $$12 \,\, \text{cm}$$ behind the lens. The new required image distance ($$v_2$$) is:

$$v_2 = 12 - \Delta x = 12 - \frac{1}{3} = \frac{35}{3} \,\, \text{cm}$$

Now, apply the thin lens formula again with our constant focal length to calculate the new object position ($$u_2$$):

$$\frac{1}{f} = \frac{1}{v_2} - \frac{1}{u_2}$$

$$\frac{21}{240} = \frac{3}{\text{35}} - \frac{1}{u_2}$$

$$\frac{1}{u_2} = \frac{3}{35} - \frac{21}{240} = \frac{3}{35} - \frac{7}{80}$$

Find a common denominator to solve for the fraction:

$$\frac{1}{u_2} = \frac{3 \times 16 - 7 \times 7}{560} = \frac{48 - 49}{560} = -\frac{1}{560}$$

$$u_2 = -560 \,\, \text{cm} = -5.6 \,\, \text{m}$$

Concept Check: Interposing the slab introduces a forward shift in the image profile. To bring the image plane back into a sharp state onto the fixed film, the source object must be pulled backwards along the principal axis to an absolute spatial position of exactly $$5.6 \,\, \text{m}$$.

Correct Option Key: Option D (5.6 m)