A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm fr

Q: Why does the glass block shift the object closer to the lens?

When light rays from the object pass through the glass block, they refract and bend away from the normal as they exit into the air. If you trace these diverging rays backward, they appear to originate from a point slightly ahead of the actual object. This is called apparent depth!

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

Not at all! The focal length is an intrinsic property of the lens, determined by its curvature and material. The glass block only changes the *effective position* of the object.

Q: Why is the shift $x$ subtracted from the original object distance?

The shift $x$ moves the apparent position of the object *in the direction of the incident light* (towards the lens). Since the original distance was 10 cm, moving 0.5 cm closer means the new distance is $10 - 0.5 = 9.5$ cm.

Q: Can I just add the 0.5 cm shift to the final image distance?

No, you cannot! The lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ is non-linear. A small change in the object distance $u$ does not result in an equal change in the image distance $v$. You must recalculate $v$ entirely.

Q: What if the glass block was placed between the lens and the screen instead?

Great question! If placed after the lens, the converging rays would pass through the block, and the block would shift the final image further away by the exact same amount $x = 0.5$ cm. But here, it's placed before the lens, so it shifts the object, which then gets magnified by the lens's non-linear behavior.

Question

A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now, a glass block (refractive index is 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance $d$. Then, $d$ is

Options:

(A) 0

(B) 1.1 cm away from the lens

(C) 0.55 cm away from the lens

(D) 0.55 cm towards the lens

Sigma Percentile · Accepted Answer

### The Case of the Shifting Screen: A Journey Through Lenses and Glass Blocks

Imagine you are setting up an optics experiment. You place a light source exactly $10 	ext{ cm}$ in front of a convex lens, and to your delight, a perfectly sharp image forms on a screen exactly $10 	ext{ cm}$ behind the lens.

This beautiful symmetry is our first major clue. In the world of optics, when the object distance ($u$) and the image distance ($v$) are equal in magnitude, it means the object is placed exactly at twice the focal length ($2f$).

Let's prove this mathematically using the lens formula:

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

Substituting our initial values ($v = +10 	ext{ cm}$ and $u = -10 	ext{ cm}$):

$$\frac{1}{10} - \frac{1}{-10} = \frac{1}{f}$$

$$\frac{2}{10} = \frac{1}{f} \implies f = 5 	ext{ cm}$$

We now know our convex lens has a focal length of **$5 	ext{ cm}$**. Keep this locked in your mind; it is the anchor for the rest of the problem.

### The Twist: Introducing the Glass Block

Now, we introduce a twist. A glass block of thickness $t = 1.5 	ext{ cm}$ and refractive index $\mu = 1.5$ is placed right next to the light source. What does this block do?

When light rays from the source pass through the glass block, they undergo refraction at both parallel surfaces. This causes a lateral shift. To the lens, the light source appears to have magically moved closer! This phenomenon is known as **apparent shift**.

The formula for the apparent shift $x$ produced by a glass slab is:

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

Let's plug in our values:

$$x = 1.5 \left(1 - \frac{1}{1.5}ight) = 1.5 \left(1 - \frac{2}{3}ight) = 1.5 	imes \frac{1}{3} = 0.5 	ext{ cm}$$

The light source appears to have shifted **$0.5 	ext{ cm}$ towards the lens**.

### Recalculating the Image Position

Because of this apparent shift, the lens now "thinks" the object is at a new distance. The original distance was $10 	ext{ cm}$, so the new object distance $u'$ is:

$$u' = -(10 - 0.5) = -9.5 	ext{ cm}$$

With our new object distance and the focal length we found earlier ($f = 5 	ext{ cm}$), we can use the lens formula again to find exactly where the new sharp image will form ($v'$):

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

$$\frac{1}{v'} - \frac{1}{-9.5} = \frac{1}{5}$$

$$\frac{1}{v'} + \frac{1}{9.5} = \frac{1}{5}$$

Moving the terms to solve for $v'$:

$$\frac{1}{v'} = \frac{1}{5} - \frac{1}{9.5} = \frac{9.5 - 5}{5 	imes 9.5} = \frac{4.5}{47.5}$$

$$v' = \frac{47.5}{4.5} = \frac{475}{45} = \frac{95}{9} \approx 10.55 	ext{ cm}$$

### The Final Shift

The new image forms at approximately **$10.55 	ext{ cm}$** from the lens. To catch this sharp image, we must move the screen from its original position of $10 	ext{ cm}$.

The required shift $d$ is simply the difference between the new and old image distances:

$$d = v' - v = 10.55 - 10 = 0.55 	ext{ cm}$$

Since the new image distance is greater than the original, the screen must be moved **away from the lens**. Therefore, the correct answer is **$0.55 	ext{ cm}$ away from the lens**.

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Visualized Solution

$I ni t ia l S e t u p$

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$F oc a l L e n g t hV a l u e$

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$L e n s F or m u l a f or N e w I ma g e$

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$C a l c u l a t in g N e w I ma g eD i s t an ce$

$S cr ee n S hi f t$

$F ina l C o n c l u s i o n$

The Sigma Insight: Lens

The Case of the Shifting Screen

The Twist

Recalculating the Image Position

The Final Shift

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