An object 2.4 m infront of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate

Q: Why does the glass plate shift the image forward?

When light rays converging towards a point pass through a parallel-sided glass slab, they undergo lateral displacement. Because they bend towards the normal upon entering and away from the normal upon exiting, the point of convergence is pushed further away in the direction of the light propagation.

Q: Why did we subtract the shift from the film's distance to get $v'$?

The film is fixed at 12 cm. The glass plate will push the image forward by 1/3 cm. If the lens still formed the image at 12 cm, the plate would push it to 12.33 cm (behind the film). To make the final image land exactly on the film at 12 cm, the lens must aim for a point 1/3 cm *before* the film, which is $12 - 1/3 = 35/3$ cm.

Q: Does the glass plate change the focal length of the lens?

No! The focal length is an intrinsic property of the lens, determined by its radii of curvature and refractive index. The glass plate is a separate optical element that simply translates the rays after they have left the lens.

Q: Why didn't we use the lens maker's formula?

We don't have the radii of curvature or the refractive index of the lens itself. However, the initial object and image distances give us all the information we need to find the focal length using the standard thin lens formula.

Q: What if the glass plate was placed before the lens?

If the plate was placed between the object and the lens, it would shift the *apparent* position of the object closer to the lens by the same amount $S$. The lens would then act on this virtual object. The final result would be different!

Question

An object $2.4 	ext{ m}$ infront of a lens forms a sharp image on a film $12 	ext{ cm}$ behind the lens. A glass plate $1 	ext{ cm}$ 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 shifted to be in sharp focus on film?

Options:

(A) $7.2 	ext{ m}$

(B) $2.4 	ext{ m}$

(C) $3.2 	ext{ m}$

(D) $5.6 	ext{ m}$

Sigma Percentile · Accepted Answer

The problem of the shifting image is a classic in optics, blending the principles of thin lenses with the subtle, yet powerful, effect of a glass slab. It’s a beautiful exercise in tracing the journey of light through multiple optical elements.

### Analyzing the Initial Setup

Imagine the scene: a lens is perfectly focusing the light from an object onto a film. The object is standing $2.4 	ext{ m}$ (or $240 	ext{ cm}$) in front of the lens, and the film is exactly $12 	ext{ cm}$ behind it.

Before we can predict how the system will react to any changes, we must uncover the hidden identity of the lens—its focal length.

### The Master Equation: Finding the Focal Length

We summon the thin lens formula:
$$ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} $$

By substituting our known values (remembering the sign convention where the object distance $u$ is negative and the image distance $v$ is positive), we get:
$$ \frac{1}{f} = \frac{1}{12} - \frac{1}{-240} $$

$$ \frac{1}{f} = \frac{20}{240} + \frac{1}{240} = \frac{21}{240} $$

So, the focal length of our lens is $f = \frac{240}{21} 	ext{ cm}$. We'll keep it as a fraction to maintain absolute precision for the upcoming steps.

### The Glass Plate Intervenes

Now, the plot thickens. A glass plate of thickness $t = 1 	ext{ cm}$ and refractive index $\mu = 1.5$ is inserted between the lens and the film.

What does a glass slab do to converging light rays? It doesn't change their convergence angle, but it laterally displaces them, effectively shifting the point of focus further down the optical axis. The formula for this longitudinal shift $S$ is:
$$ S = t\left(1 - \frac{1}{\mu}ight) $$

Plugging in our values:
$$ S = 1 	imes \left(1 - \frac{1}{1.5}ight) = 1 	imes \left(1 - \frac{2}{3}ight) = \frac{1}{3} 	ext{ cm} $$

The glass plate pushes the image forward by exactly $\frac{1}{3} 	ext{ cm}$.

### The Final Calculation: Shifting the Object

Here is the crucial logical leap. We want the final image to land *exactly* on the film, which is fixed at $12 	ext{ cm}$. Because the glass plate will push the image forward by $\frac{1}{3} 	ext{ cm}$, the lens must now aim to form the image $\frac{1}{3} 	ext{ cm}$ *before* the film.

Our new target image distance for the lens is:
$$ v' = 12 - \frac{1}{3} = \frac{35}{3} 	ext{ cm} $$

To force the lens to form the image at this new $v'$, we must move the object to a new position, $u'$. We call upon the lens formula one last time:
$$ \frac{1}{u'} = \frac{1}{v'} - \frac{1}{f} $$

Substitute the values we've gathered:
$$ \frac{1}{u'} = \frac{3}{35} - \frac{21}{240} $$

To subtract these fractions, we find a common denominator. The least common multiple of $35$ and $240$ is $560$.
$$ \frac{1}{u'} = \frac{48}{560} - \frac{49}{560} = -\frac{1}{560} $$

Inverting this gives us the new object distance:
$$ u' = -560 	ext{ cm} = -5.6 	ext{ m} $$

**Conclusion:** To maintain a sharp focus on the film after the glass plate is introduced, the object must be shifted to a distance of **$5.6 	ext{ m}$** from the lens.

Select Answer:

Visualized Solution

Initial Setup

Focal Length of the Lens

Calculating Focal Length

Introducing the Glass Plate

Shift Produced by Glass Plate

New Image Position for Lens

Finding the New Object Position

Final Calculation

Conclusion

The Sigma Insight: Lens

Analyzing the Initial Setup

The Master Equation

The Glass Plate Intervenes

The Final Calculation

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