Given:

• Medium 1 (air): $n_1 = 1.00$
• Medium 2 (water): $n_2 = 1.33$
• Incident angle: $\theta_1 = 30°$

Solution:

Apply Snell's Law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$

Check: Entering denser medium (n₂ > n₁), so bends toward normal. ✓ θ₂ = 22.1° < θ₁ = 30° ✓

Answer: θ₂ = 22.1° (bends toward normal)

Given:

• Medium 1 (air): $n_1 = 1.00$
• Medium 2 (water): $n_2 = 1.33$
• Incident angle: $\theta_1 = 30°$

Solution:

Apply Snell's Law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$

Check: Entering denser medium (n₂ > n₁), so bends toward normal. ✓ θ₂ = 22.1° < θ₁ = 30° ✓

Answer: θ₂ = 22.1° (bends toward normal)

Given:

• Medium 1 (air): $n_1 = 1.00$
• Medium 2 (water): $n_2 = 1.33$
• Incident angle: $\theta_1 = 30°$

Solution:

Apply Snell's Law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$

Check: Entering denser medium (n₂ > n₁), so bends toward normal. ✓ θ₂ = 22.1° < θ₁ = 30° ✓

Answer: θ₂ = 22.1° (bends toward normal)

Given:

• Medium 1 (air): $n_1 = 1.00$
• Medium 2 (water): $n_2 = 1.33$
• Incident angle: $\theta_1 = 30°$

Solution:

Apply Snell's Law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$

Check: Entering denser medium (n₂ > n₁), so bends toward normal. ✓ θ₂ = 22.1° < θ₁ = 30° ✓

Answer: θ₂ = 22.1° (bends toward normal)

Given:

• Medium 1 (glass): $n_1 = 1.50$
• Medium 2 (air): $n_2 = 1.00$

Part 1: Critical angle

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667$

Part 2: At 45° incidence

Since $45° > \theta_c = 41.8°$, we have total internal reflection!

Light reflects completely back into glass. No refraction into air.

Answer:

• Critical angle: θ_c = 41.8°
• At 45°: Total internal reflection (no light exits to air)

This is why fiber optic cables work - light trapped inside!

Given:

• Medium 1 (glass): $n_1 = 1.50$
• Medium 2 (air): $n_2 = 1.00$

Part 1: Critical angle

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667$

Part 2: At 45° incidence

Since $45° > \theta_c = 41.8°$, we have total internal reflection!

Light reflects completely back into glass. No refraction into air.

Answer:

• Critical angle: θ_c = 41.8°
• At 45°: Total internal reflection (no light exits to air)

This is why fiber optic cables work - light trapped inside!

Given:

• Medium 1 (glass): $n_1 = 1.50$
• Medium 2 (air): $n_2 = 1.00$

Part 1: Critical angle

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667$

Part 2: At 45° incidence

Since $45° > \theta_c = 41.8°$, we have total internal reflection!

Light reflects completely back into glass. No refraction into air.

Answer:

• Critical angle: θ_c = 41.8°
• At 45°: Total internal reflection (no light exits to air)

This is why fiber optic cables work - light trapped inside!

Given:

• Medium 1 (glass): $n_1 = 1.50$
• Medium 2 (air): $n_2 = 1.00$

Part 1: Critical angle

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667$

Part 2: At 45° incidence

Since $45° > \theta_c = 41.8°$, we have total internal reflection!

Light reflects completely back into glass. No refraction into air.

Answer:

• Critical angle: θ_c = 41.8°
• At 45°: Total internal reflection (no light exits to air)

This is why fiber optic cables work - light trapped inside!

Given:

• Medium 1 (glass): $n_1 = 1.50$
• Medium 2 (air): $n_2 = 1.00$

Part 1: Critical angle

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667$

Part 2: At 45° incidence

Since $45° > \theta_c = 41.8°$, we have total internal reflection!

Light reflects completely back into glass. No refraction into air.

Answer:

• Critical angle: θ_c = 41.8°
• At 45°: Total internal reflection (no light exits to air)

This is why fiber optic cables work - light trapped inside!

Given:

• Focal length: $f = 20$ cm (positive for concave)
• Object distance: $d_o = 60$ cm

Part (a): Image distance

Mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

Part (b): Magnification

$m = -\frac{d_i}{d_o} = -\frac{30}{60} = -0.50$

Part (c): Image description

• $d_i > 0$: Real image (in front of mirror)
• $m < 0$: Inverted
• $|m| < 1$: (half size)

Answer:

• (a) d_i = 30 cm (in front)
• (b) m = -0.50
• (c) Real, inverted, reduced to half size

Given:

• Focal length: $f = 20$ cm (positive for concave)
• Object distance: $d_o = 60$ cm

Part (a): Image distance

Mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

Part (b): Magnification

$m = -\frac{d_i}{d_o} = -\frac{30}{60} = -0.50$

Part (c): Image description

• $d_i > 0$: Real image (in front of mirror)
• $m < 0$: Inverted
• $|m| < 1$: (half size)

Answer:

• (a) d_i = 30 cm (in front)
• (b) m = -0.50
• (c) Real, inverted, reduced to half size

Given:

• Focal length: $f = 20$ cm (positive for concave)
• Object distance: $d_o = 60$ cm

Part (a): Image distance

Mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

Part (b): Magnification

$m = -\frac{d_i}{d_o} = -\frac{30}{60} = -0.50$

Part (c): Image description

• $d_i > 0$: Real image (in front of mirror)
• $m < 0$: Inverted
• $|m| < 1$: (half size)

Answer:

• (a) d_i = 30 cm (in front)
• (b) m = -0.50
• (c) Real, inverted, reduced to half size

Given:

• Focal length: $f = 20$ cm (positive for concave)
• Object distance: $d_o = 60$ cm

Part (a): Image distance

Mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

Part (b): Magnification

$m = -\frac{d_i}{d_o} = -\frac{30}{60} = -0.50$

Part (c): Image description

• $d_i > 0$: Real image (in front of mirror)
• $m < 0$: Inverted
• $|m| < 1$: (half size)

Answer:

• (a) d_i = 30 cm (in front)
• (b) m = -0.50
• (c) Real, inverted, reduced to half size

Given:

• Focal length: $f = 20$ cm (positive for concave)
• Object distance: $d_o = 60$ cm

Part (a): Image distance

Mirror equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

Part (b): Magnification

$m = -\frac{d_i}{d_o} = -\frac{30}{60} = -0.50$

Part (c): Image description

• $d_i > 0$: Real image (in front of mirror)
• $m < 0$: Inverted
• $|m| < 1$: (half size)

Answer:

• (a) d_i = 30 cm (in front)
• (b) m = -0.50
• (c) Real, inverted, reduced to half size