Refraction of Light

Given: n = 1.50, c = 3 × 10⁸ m/s

1. Substitute: v = 3 × 10⁸ / 1.50
2. Calculate: v = 2.00e+08 m/s

Given: n₁ = 1.33, n₂ = 1.00, i = 30°

1. sin r = (1.33 sin 30°)/1.00
2. sin r = 0.665
3. r = 41.68°

Given: real depth = 6 m, n = 1.5

1. Apparent depth = 6/1.5
2. Apparent depth = 4.00 m

Given: t = 5 cm, i = 40°, r = 25°

1. d = 5 sin(15°)/cos(25°)
2. d = 1.43 cm

Given: n₁=1.33, n₂=1.50

1. n₂₁ = 1.50/1.33
2. n₂₁ = 1.13

Given: n = 1.50, c = 3 × 10⁸ m/s

1. Substitute: v = 3 × 10⁸ / 1.50
2. Calculate: v = 2.00e+08 m/s

Given: n₁ = 1.00, n₂ = 1.33, i = 45°

1. sin r = (1.00 sin 45°)/1.33
2. sin r = 0.532
3. r = 32.12°

Given: n = 1.60, c = 3 × 10⁸ m/s

1. Substitute: v = 3 × 10⁸ / 1.60
2. Calculate: v = 1.88e+08 m/s

Given: real depth = 6 m, n = 1.5

1. Apparent depth = 6/1.5
2. Apparent depth = 4.00 m

Given: n₁=1.52, n₂=1.33

1. n₂₁ = 1.33/1.52
2. n₂₁ = 0.88

Given: t = 6 cm, i = 45°, r = 28°

1. d = 6 sin(17°)/cos(28°)
2. d = 1.99 cm

Given: n₁ = 1.00, n₂ = 1.33, i = 45°

1. sin r = (1.00 sin 45°)/1.33
2. sin r = 0.532
3. r = 32.12°

Given: n = 1.60, c = 3 × 10⁸ m/s

1. Substitute: v = 3 × 10⁸ / 1.60
2. Calculate: v = 1.88e+08 m/s

Given: n₁ = 1.00, n₂ = 1.33, i = 45°

1. sin r = (1.00 sin 45°)/1.33
2. sin r = 0.532
3. r = 32.12°

Given: t = 4 cm, i = 35°, r = 22°

1. d = 4 sin(13°)/cos(22°)
2. d = 0.97 cm

Given: n₁=1.33, n₂=2.42

1. n₂₁ = 2.42/1.33
2. n₂₁ = 1.82

Given: n = 2.42, c = 3 × 10⁸ m/s

1. Substitute: v = 3 × 10⁸ / 2.42
2. Calculate: v = 1.24e+08 m/s

Given: real depth = 3 m, n = 1.33

1. Apparent depth = 3/1.33
2. Apparent depth = 2.26 m

Given: n₁ = 1.00, n₂ = 1.33, i = 45°

1. sin r = (1.00 sin 45°)/1.33
2. sin r = 0.532
3. r = 32.12°

Given: t = 4 cm, i = 35°, r = 22°

1. d = 4 sin(13°)/cos(22°)
2. d = 0.97 cm

Given: n₁ = 1.00, n₂ = 1.33, i = 45°

1. sin r = (1.00 sin 45°)/1.33
2. sin r = 0.532
3. r = 32.12°

Given: t = 4 cm, i = 35°, r = 22°

1. d = 4 sin(13°)/cos(22°)
2. d = 0.97 cm

Given: n₁=1.33, n₂=2.42

1. n₂₁ = 2.42/1.33
2. n₂₁ = 1.82

Given: real depth = 1.8 m, n = 1.33

1. Apparent depth = 1.8/1.33
2. Apparent depth = 1.35 m

Given: n = 1.33, c = 3 × 10⁸ m/s

1. Substitute: v = 3 × 10⁸ / 1.33
2. Calculate: v = 2.26e+08 m/s

Given: n₁ = 1.00, n₂ = 1.33, i = 45°

1. sin r = (1.00 sin 45°)/1.33
2. sin r = 0.532
3. r = 32.12°

Given: n = 1.50, c = 3 × 10⁸ m/s

1. Substitute: v = 3 × 10⁸ / 1.50
2. Calculate: v = 2.00e+08 m/s

Given: real depth = 2 m, n = 1.33

1. Apparent depth = 2/1.33
2. Apparent depth = 1.50 m

Given: n₁ = 1.50, n₂ = 1.00, i = 30°

1. sin r = (1.50 sin 30°)/1.00
2. sin r = 0.750
3. r = 48.59°

Given: n₁=1.52, n₂=1.33

1. n₂₁ = 1.33/1.52
2. n₂₁ = 0.88

Given: real depth = 4.5 m, n = 1.5

1. Apparent depth = 4.5/1.5
2. Apparent depth = 3.00 m

Given: n = 1.60, c = 3 × 10⁸ m/s

1. Substitute: v = 3 × 10⁸ / 1.60
2. Calculate: v = 1.88e+08 m/s

Given: n₁ = 1.50, n₂ = 1.00, i = 30°

1. sin r = (1.50 sin 30°)/1.00
2. sin r = 0.750
3. r = 48.59°

Given: t = 5 cm, i = 40°, r = 25°

1. d = 5 sin(15°)/cos(25°)
2. d = 1.43 cm

Given: real depth = 3 m, n = 1.33

1. Apparent depth = 3/1.33
2. Apparent depth = 2.26 m

Given: n₁ = 1.00, n₂ = 1.33, i = 45°

1. sin r = (1.00 sin 45°)/1.33
2. sin r = 0.532
3. r = 32.12°

Given: n₁=1.50, n₂=1.00

1. n₂₁ = 1.00/1.50
2. n₂₁ = 0.67

Given: real depth = 6 m, n = 1.5

1. Apparent depth = 6/1.5
2. Apparent depth = 4.00 m

Given: t = 5 cm, i = 40°, r = 25°

1. d = 5 sin(15°)/cos(25°)
2. d = 1.43 cm

Given: n₁ = 1.00, n₂ = 1.50, i = 30°

1. sin r = (1.00 sin 30°)/1.50
2. sin r = 0.333
3. r = 19.47°

Given: n = 1.50, c = 3 × 10⁸ m/s

1. Substitute: v = 3 × 10⁸ / 1.50
2. Calculate: v = 2.00e+08 m/s

Given: n₁ = 1.33, n₂ = 1.00, i = 30°

1. sin r = (1.33 sin 30°)/1.00
2. sin r = 0.665
3. r = 41.68°

Given: real depth = 6 m, n = 1.5

1. Apparent depth = 6/1.5
2. Apparent depth = 4.00 m

Given: n₁=1.52, n₂=1.33

1. n₂₁ = 1.33/1.52
2. n₂₁ = 0.88

Given: t = 6 cm, i = 45°, r = 28°

1. d = 6 sin(17°)/cos(28°)
2. d = 1.99 cm

Because light slows down in glass; the part of the wavefront entering first slows earlier, so the ray bends towards the normal.

Speed decreases because n = c/v.

The two slab faces are parallel, so the refraction at the second face reverses the angular deviation of the first face.

It depends on slab thickness, refractive index and angle of incidence.

Rays bend away from the normal while coming from water to air, so backward extensions meet at a shallower point.

Because light slows down in glass; the part of the wavefront entering first slows earlier, so the ray bends towards the normal.

Speed decreases because n = c/v.

The two slab faces are parallel, so the refraction at the second face reverses the angular deviation of the first face.

It depends on slab thickness, refractive index and angle of incidence.

Rays bend away from the normal while coming from water to air, so backward extensions meet at a shallower point.

Because light slows down in glass; the part of the wavefront entering first slows earlier, so the ray bends towards the normal.

Speed decreases because n = c/v.

The two slab faces are parallel, so the refraction at the second face reverses the angular deviation of the first face.

It depends on slab thickness, refractive index and angle of incidence.

Rays bend away from the normal while coming from water to air, so backward extensions meet at a shallower point.

Because light slows down in glass; the part of the wavefront entering first slows earlier, so the ray bends towards the normal.

Speed decreases because n = c/v.

The two slab faces are parallel, so the refraction at the second face reverses the angular deviation of the first face.

It depends on slab thickness, refractive index and angle of incidence.

Rays bend away from the normal while coming from water to air, so backward extensions meet at a shallower point.

Because light slows down in glass; the part of the wavefront entering first slows earlier, so the ray bends towards the normal.

Speed decreases because n = c/v.

The two slab faces are parallel, so the refraction at the second face reverses the angular deviation of the first face.

It depends on slab thickness, refractive index and angle of incidence.

Rays bend away from the normal while coming from water to air, so backward extensions meet at a shallower point.

Because light slows down in glass; the part of the wavefront entering first slows earlier, so the ray bends towards the normal.

Speed decreases because n = c/v.

The two slab faces are parallel, so the refraction at the second face reverses the angular deviation of the first face.

It depends on slab thickness, refractive index and angle of incidence.

Rays bend away from the normal while coming from water to air, so backward extensions meet at a shallower point.

Because light slows down in glass; the part of the wavefront entering first slows earlier, so the ray bends towards the normal.

Speed decreases because n = c/v.

The two slab faces are parallel, so the refraction at the second face reverses the angular deviation of the first face.

It depends on slab thickness, refractive index and angle of incidence.

Rays bend away from the normal while coming from water to air, so backward extensions meet at a shallower point.

Light slows down in glass. Due to the change in speed at the boundary, the direction changes towards the normal.

Yes. If light falls normally on the interface, its speed changes but its direction does not change.

Glass has higher refractive index than air, so light travels slower in glass.

The first surface bends the ray, and the second parallel surface bends it back parallel to original direction, leaving a sideways displacement.

Rays from the bottom bend away from the normal while emerging, and their backward extensions meet above the real bottom.

Yes. Different colours have different speeds in a medium, so refractive index depends on wavelength.

Both statements are true. The reason correctly explains why the coin appears raised.

The reason is true, but the assertion is false. At normal incidence direction does not change.

Both statements are true and the reason correctly explains the assertion.

Both statements are true and the reason correctly explains the assertion.

Both statements are true and the reason correctly explains the trend.

Correct matching:

1. Snell’s law → n₁ sin i = n₂ sin r
2. Absolute refractive index → c/v
3. Glass slab → Lateral shift
4. Apparent depth → real depth/n
5. Optical density → higher refractive index

1. Air has lowest refractive index.
2. Water has n ≈ 1.33.
3. Glass has n ≈ 1.5.
4. Diamond has n ≈ 2.42.

Identify each formula by its defining quantity: refractive index, lateral shift, relative refractive index and apparent depth.

Identify each formula by its defining quantity: refractive index, lateral shift, relative refractive index and apparent depth.

Identify each formula by its defining quantity: refractive index, lateral shift, relative refractive index and apparent depth.

Yes. If incidence is normal, the ray direction remains unchanged even though speed changes.

The slab causing a larger difference between i and r produces more lateral shift. Usually higher refractive index gives larger shift.

It assumes near-normal viewing. For large angles, exact ray tracing is required.

Yes. If medium 2 is optically rarer than medium 1, n₂₁ = n₂/n₁ is less than 1.

Atmospheric layers have different refractive indices, so light bends gradually.

Statement I is true. Statement II is false because light speed changes in media.

Statement I is true. Statement II is false; there is lateral shift for oblique incidence.

Both statements are true.

Statement I is true. Statement II is false for standard Snell substitution.

1. It occurs due to change in speed.
2. It follows Snell’s law.
3. At normal incidence, direction may remain unchanged.
4. It cannot occur in transparent media is false.

Lateral shift depends on thickness, angle of incidence and refractive index of slab.

Absolute refractive index, relative refractive index and sin i / sin r are ratios, so they are dimensionless.

Light speed is lower and refractive index is higher. Wavelength decreases, while frequency remains same.

v = 3 × 10⁸ / 1.5 = 2 × 10⁸ m/s.

x = 6/1.5 = 4.

x = 2/1 = 2.

x = 9/6 = 1.5, nearest integer is 2 if integer rounding is required.

At each boundary Snell’s law applies. Therefore the quantity n sin θ remains constant across all layers.

For the same i and r, d is directly proportional to t.

The boundary cannot create or destroy wave cycles, so frequency is fixed by the source. Speed and wavelength change.

If the required sin r exceeds 1, no refracted ray exists and total internal reflection occurs.

The eye traces refracted rays backwards in straight lines; those extensions meet above the real object, so the object appears raised.