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Refraction at Plane Surfaces

Class 10 Physics • Chapter 04 (Deep Detail)

1. Refraction of Light

Definition: The bending of light when it passes from one transparent medium to another is called Refraction.

Cause: Change in speed of light in different media.

Rules of Refraction:

SNELL'S LAW

1. The incident ray, the refracted ray, and the normal at the point of incidence, all lie in the same plane.

2. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media.

$$ \frac{\sin i}{\sin r} = \mu \text{ (Refractive Index)} $$
Topic Check: Snell's Law

Q: Light travels from air to water. If $\mu_w = 1.33$, will it bend towards or away from normal?

Ans: Towards. (Rarer to Denser).

Practice Q1: Bending Logic
[IMAGE PLACEHOLDER: SNELL LAW REFRACTION]
Ray traveling from Medium A (n=1.5) to Medium B (n=1.2). Show ray bending AWAY from normal. Mark angle i and r. (Invert in dark mode).

REASONING A ray of light passes from glass ($\mu=1.5$) to water ($\mu=1.33$). Will it bend towards or away from the normal? Calculate the ratio $\frac{\sin i}{\sin r}$.

Solution: Glass to Water means Denser to Rarer. Ray bends Away from normal.

Refractive index of water w.r.t glass = $_g\mu_w$

$= \frac{\mu_w}{\mu_g} = \frac{1.33}{1.5} \approx 0.89$.

So, $\frac{\sin i}{\sin r} = 0.89$.

Practice Q2: Wave Properties

CONCEPTUAL When light travels from air to glass, which property of the wave remains unchanged?

Ans: Frequency remains unchanged. Speed and Wavelength decrease.

2. Refractive Index ($\mu$)

It is a measure of how much a medium slows down light.

$$ \mu = \frac{\text{Speed in vacuum (c)}}{\text{Speed in medium (v)}} $$

Speed of light in vacuum $c = 3 \times 10^8$ m/s.

Examples: $\mu_{glass} = 1.5$, $\mu_{water} = 1.33$.

Practice Q3: Calculation

NUMERICAL The speed of light in air is $3 \times 10^8$ m/s and in glass is $2 \times 10^8$ m/s. Calculate refractive index.

Solution: $\mu = c/v = 3/2 = 1.5$.

CONCEPT CHECK Higher Refractive Index $\implies$ Denser Optical Medium $\implies$ Slower Speed of Light $\implies$ More Bending.

3. Refraction Through a Glass Block

[IMAGE PLACEHOLDER: GLASS SLAB REFRACTION]
Diagram of refraction through a glass slab. Label Incident Ray, Refracted Ray, Emergent Ray, Angle i, r, e. Show Emergent ray parallel to incident ray. Mark Lateral Displacement.

When a ray passes through a rectangular glass slab:

  1. Ray Refracts at first surface (Air $\to$ Glass): Bends towards normal.
  2. Ray Refracts at second surface (Glass $\to$ Air): Bends away from normal.
  3. Result: Emergent Ray is parallel to Incident Ray but shifted sideways.

Lateral Displacement

The perpendicular distance between the incident ray produced forward and the emergent ray.

Factors increasing Lateral Displacement:

Practice Q4: Diagram Check
[IMAGE PLACEHOLDER: GLASS SLAB LATERAL DISPLACEMENT]
Rectangular glass slab. Incident ray at angle 'i'. Refracted ray bends towards normal 'r'. Emergent ray 'e' is parallel to incident. Mark 'Lateral Displacement' x. (Invert in dark mode).

BOARD CHECK Draw a ray diagram to show the refraction of a monochromatic ray through a parallel sided glass block. Show that incident ray is parallel to emergent ray.

Verification: Apply Snell's Law at both surfaces. Since surfaces are parallel, angle of refraction at first surface equals angle of incidence at second surface ($r_1 = r_2$). Thus, final emergence angle $e = i$.

Practice Q5: Lateral Displacement Factors

THINKING Does the lateral displacement produced by a glass slab depend on the thickness of the slab? If yes, how?

Solution: Yes. Directly proportional. Thicker slab $\implies$ More shift.

4. Real and Apparent Depth

An object placed in a denser medium (like water) appears raised when viewed from a rarer medium (air).

Shift: The distance by which the object appears raised.

$$ \text{Shift} = \text{Real Depth} \times \left(1 - \frac{1}{\mu}\right) $$
Practice Q6: Microscope Problem

NUMERICAL A microscope is focused on a mark on a paper. A glass slab of thickness 3 cm and refractive index 1.5 is placed over the mark. By how much amount should the microscope be moved to focus the mark again?

Solution: The image will shift upwards by:

$\text{Shift} = \text{Real Thickness} \times (1 - 1/\mu)$

$S = 3 \times (1 - 1/1.5) = 3 \times (1 - 2/3)$

$= 3 \times (1/3) = 1 \text{ cm}$.

Ans: Move microscope up by 1 cm.

Practice Q7: Coin in Water

NUMERICAL A coin placed at the bottom of a beaker appears to be raised by 4 cm. If the refractive index of water is $4/3$, find the depth of the water in the beaker.

Solution: Shift=4. Formula $S = \text{Real} \times (1 - 1/\mu)$.

$4 = x(1 - 3/4) = x(1/4)$. Therefore $x = 16 \text{ cm}$.

5. Prism

A transparent medium bounded by two plane surfaces inclined at an angle (Angle of Prism $A$).

DEVIATION FORMULA
$$ i + e = A + \delta $$

Where $i$ = angle of incidence, $e$ = angle of emergence, $A$ = Angle of prism, $\delta$ = Angle of deviation.

Practice Q8: Deviation Factors

CONCEPTUAL Which color of white light deviates the most and which one least when passing through a prism?

Ans: Violet deviates most (highest $\mu$). Red deviates least (lowest $\mu$).

6. Total Internal Reflection (TIR)

[IMAGE PLACEHOLDER: TOTAL INTERNAL REFLECTION]
Three diagrams showing rays from denser to rarer: 1. Refraction (i < C). 2. Grazing emergence (i=C). 3. Total Internal Reflection (i> C).

When a ray travels from Denser to Rarer medium:

Critical Angle ($C$): The angle of incidence in denser medium for which the angle of refraction in rarer medium is $90^\circ$.

Conditions for TIR:

  1. Light must travel from Denser to Rarer medium.
  2. Angle of incidence must be greater than Critical Angle ($i > C$).

For Glass ($ \mu = 1.5 $), $C \approx 42^\circ$.

Practice Q9: Prism Path (TIR)
[IMAGE PLACEHOLDER: PRISM TIR 90 DEGREE]
A total reflecting prism (Isosceles Right Angled Prism ABC, B=90). A ray enters normally on face AB. Show path: hits AC at 45 deg, TIR occurs (since 45 > 42), reflects and exits normally from BC. (Invert in dark mode).

DIAGRAM Complete the path of the ray entering the prism normally. ($C = 42^\circ$).

Solution:

1. Enters AB undeflected ($i=0$).

2. Hits hypotenuse AC at $45^\circ$.

3. Since $45^\circ > 42^\circ$ (Critical angle), TIR takes place.

4. Ray deviates by $90^\circ$ and emerges from BC.

Practice Q10: Right Angled Prism

CONCEPTUAL A ray of light is incident normally on one of the faces of a right-angled isosceles prism. Draw the path of the ray. (Take $\mu = 1.5$)

Solution: See diagram above. Ray enters normally, TIR at hypotenuse ($45^\circ > 42^\circ$), emerges normally from other leg.

BOARD EXAM FAVORITE Total Reflecting Prisms: Right-angled isosceles prisms ($45^\circ$-$90^\circ$-$45^\circ$) are used to deviate light by $90^\circ$ or $180^\circ$ using TIR. Practice these ray diagrams thoroughly (Periscope, Binoculars).

Factors Affecting Angle of Deviation (Prism)