Real and Apparent Depth: Glass Slab Shift | Class 10 Light

Quick answer: An object under glass or water looks raised because light bends away from the normal as it leaves the denser medium. When viewed from almost directly above, apparent depth = real depth ÷ n, and the normal shift is $s = t\left(1 - \frac{1}{n}\right)$. For a 9 cm glass slab with $n = 1.5$, the letters appear 6 cm deep, so they seem raised by 3 cm.

Syllabus note: The rationalised NCERT Class 10 chapter explains this effect in words (a coin in water or letters under a glass slab look raised). The formulas for apparent depth and normal shift are taught in Class 12 (Ray Optics). Use the numbers below for understanding and practice, and check your syllabus before relying on them in a Class 10 exam.

Real depth and apparent depth

• Real depth: the actual distance of an object below the surface.
• Apparent depth: the depth at which the object appears to be when seen from above.

Light from an object in glass or water bends away from the normal as it passes into air. Our eyes trace the rays back in straight lines, so the object appears to be at a point above its true position.

      eye (in air)
         \
          \   ray bends away
           \  from the normal
 ~~~~~~~~~~~~~~~~~~~~~~ surface
          |
          * I  image (apparent)
          |
          * O  object (real)

When the object is viewed nearly straight from above, Snell's law gives:

$n = \frac{\text{real depth}}{\text{apparent depth}}$

So:

$\text{apparent depth} = \frac{\text{real depth}}{n}$

Since $n > 1$ for glass and water, the apparent depth is less than the real depth.

Normal shift

The normal shift $s$ is the distance through which the object appears to be raised.

Step 1: Shift = real depth − apparent depth.

$s = t - \frac{t}{n}$

Step 2: Take $t$ common.

$s = t\left(1 - \frac{1}{n}\right)$

Here $t$ is the thickness of the slab (or depth of the liquid) and $n$ is its refractive index.

What does the shift depend on?

1. Thickness: a thicker slab gives a larger shift ($s \propto t$).
2. Refractive index: a larger $n$ gives a larger shift.
3. Colour of light: glass has a slightly larger refractive index for violet light than for red, so violet letters appear raised slightly more than red ones.
4. For near-vertical viewing, the shift does not depend on how high the eye is above the slab.

Solved example: letters under a glass slab

A glass slab 9 cm thick with refractive index 1.5 is placed over a printed page. By how much do the letters appear to be raised?

Step 1: Given. $t = 9$ cm, $n = 1.5$.

Step 2: Apparent depth.

$\text{apparent depth} = \frac{9}{1.5} = 6 \text{ cm}$

Step 3: Shift.

$s = 9 - 6 = 3 \text{ cm}$

Check with the formula:

$s = 9\left(1 - \frac{1}{1.5}\right) = 9 \times \frac{1}{3} = 3 \text{ cm}$

Answer: The letters appear to be raised by 3 cm. They seem to be 6 cm below the top surface of the slab.

Several layers

If an object is under several layers of different media, the total shift is the sum of the shifts in each layer:

$s = t_1\left(1 - \frac{1}{n_1}\right) + t_2\left(1 - \frac{1}{n_2}\right)$

Example: A coin lies under 4 cm of water ($n = 4/3$) with 3 cm of glass ($n = 3/2$) on top. Shift in water = 4 × 1/4 = 1 cm. Shift in glass = 3 × 1/3 = 1 cm. Total shift = 2 cm.

Summary table

Exam tips

• Keep the two answers separate: apparent depth (6 cm here) is where the object seems to be; shift (3 cm here) is how much it seems to rise.
• Write $n = 1.5$ as $\frac{3}{2}$ to make the division easy.
• In word answers, explain that light bends away from the normal when going from glass or water into air.
• Remember the result is for viewing from almost directly above.

Common mistakes

• Writing n = apparent depth ÷ real depth. For an object in a denser medium seen from air, n = real ÷ apparent, which is greater than 1.
• Giving the apparent depth when the question asks for the shift (or the other way round).
• Saying light bends towards the normal when leaving glass. It bends away from the normal.
• Using the slab thickness as the shift.

Frequently Asked Questions

Why does a coin in water appear raised?

Light from the coin bends away from the normal as it leaves water. The eye traces the rays back in straight lines, so the coin appears to be at a shallower depth than it really is.

What is the formula for apparent depth?

Apparent depth = real depth ÷ n, where n is the refractive index of the medium, for viewing from almost directly above.

What is the normal shift formula?

The normal shift is s = t(1 − 1/n), where t is the real depth or thickness and n is the refractive index.

By how much do letters under a 9 cm glass slab of refractive index 1.5 appear raised?

They appear raised by 3 cm, because the apparent depth is 9 ÷ 1.5 = 6 cm and 9 − 6 = 3 cm.

Is the apparent depth formula in the Class 10 syllabus?

The rationalised NCERT Class 10 book explains the raised appearance in words, but the formulas for apparent depth and shift are covered in Class 12 Physics.