Solution to Question 14
1. Understanding the System
The person is wearing plano-concave glasses with the flat surface facing the eyes (inside) and the concave surface facing outside.
Case 1 (Reflection): The flower is $5.0$ m behind the person. Light enters from behind, hits the flat surface first, then the concave surface. Two images are seen in front. This implies the lens is acting as a mirror (due to partial reflections).
Case 2 (Refraction): The person turns around and looks through the glasses. The flower is $5.0$ m in front, and the image is $2.5$ m in front.
2. Analyzing Case 2: Standard Refraction
Object distance $u = -5.0$ m. Virtual image distance $v = -2.5$ m.
Using the lens formula:
$$ \frac{1}{v} – \frac{1}{u} = \frac{1}{f} $$
$$ \frac{1}{-2.5} – \frac{1}{-5} = \frac{1}{f} \implies -\frac{2}{5} + \frac{1}{5} = \frac{1}{f} \implies f = -5 \text{ m} $$
For a plano-concave lens with flat surface $R_2 = \infty$ and concave surface $R_1 = -R$ (facing incoming light in this orientation, light hits concave first when looking through).
Using the Lens Maker’s equation for a plano-concave lens ($f$ is negative):
$$ \frac{1}{|f|} = \frac{\mu – 1}{R} \implies \frac{1}{5} = \frac{\mu – 1}{R} \implies R = 5(\mu – 1) \quad \text{…(i)} $$
3. Analyzing Case 1: Reflection from Behind
Light comes from the flower ($u = -5$ m), enters the flat back surface.
Image 1 (5.0 m in front):
This corresponds to reflection from the first surface (the flat surface). A flat glass surface acts as a weak plane mirror.
For a plane mirror, object at distance 5 m forms a virtual image at distance 5 m inside the mirror (which looks like 5 m “in front” of the person).
This explains the first image.
Image 2 (0.714 m in front):
This arises from light refracting into the glass, reflecting off the second surface (concave), and refracting back out.
Step A: Refraction at Flat Surface (Air $\to$ Glass)
$$ v_1 = \mu u = -5\mu $$
Step B: Reflection at Curved Surface (Glass $\to$ Air interface)
The surface is concave from the outside, so from the inside (glass side), it is convex.
Focal length of this mirror $f_m = +R/2$.
Object distance for mirror $u_2 = v_1 = -5\mu$.
Mirror formula:
$$ \frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f_m} \implies \frac{1}{v_2} – \frac{1}{5\mu} = \frac{2}{R} $$
$$ \frac{1}{v_2} = \frac{2}{R} + \frac{1}{5\mu} = \frac{10\mu + R}{5\mu R} \implies v_2 = \frac{5\mu R}{10\mu + R} $$
Step C: Refraction at Flat Surface (Glass $\to$ Air)
Apparent depth formula: $v_{\text{final}} = v_2 / \mu$.
$$ v_{\text{final}} = \frac{5R}{10\mu + R} $$
Given $v_{\text{final}} = 0.714 \approx 5/7$ m:
$$ \frac{5R}{10\mu + R} = \frac{5}{7} \implies 7R = 10\mu + R \implies 6R = 10\mu \implies R = \frac{5\mu}{3} \quad \text{…(ii)} $$
4. Calculation
Equating the two expressions for $R$ from (i) and (ii):
$$ 5(\mu – 1) = \frac{5\mu}{3} $$
$$ \mu – 1 = \frac{\mu}{3} $$
$$ \mu – \frac{\mu}{3} = 1 $$
$$ \frac{2\mu}{3} = 1 \implies \mu = 1.5 $$
$$ \mu = 1.5 $$
1. Understanding the System
The person is wearing plano-concave glasses with the flat surface facing the eyes (inside) and the concave surface facing outside.
Case 1 (Reflection): The flower is $5.0$ m behind the person. Light enters from behind, hits the flat surface first, then the concave surface. Two images are seen in front. This implies the lens is acting as a mirror (due to partial reflections).
Case 2 (Refraction): The person turns around and looks through the glasses. The flower is $5.0$ m in front, and the image is $2.5$ m in front.
2. Analyzing Case 2: Standard Refraction
Object distance $u = -5.0$ m. Virtual image distance $v = -2.5$ m. Using the lens formula: $$ \frac{1}{v} – \frac{1}{u} = \frac{1}{f} $$ $$ \frac{1}{-2.5} – \frac{1}{-5} = \frac{1}{f} \implies -\frac{2}{5} + \frac{1}{5} = \frac{1}{f} \implies f = -5 \text{ m} $$
For a plano-concave lens with flat surface $R_2 = \infty$ and concave surface $R_1 = -R$ (facing incoming light in this orientation, light hits concave first when looking through).
Using the Lens Maker’s equation for a plano-concave lens ($f$ is negative):
$$ \frac{1}{|f|} = \frac{\mu – 1}{R} \implies \frac{1}{5} = \frac{\mu – 1}{R} \implies R = 5(\mu – 1) \quad \text{…(i)} $$
3. Analyzing Case 1: Reflection from Behind
Light comes from the flower ($u = -5$ m), enters the flat back surface.
Image 1 (5.0 m in front): This corresponds to reflection from the first surface (the flat surface). A flat glass surface acts as a weak plane mirror. For a plane mirror, object at distance 5 m forms a virtual image at distance 5 m inside the mirror (which looks like 5 m “in front” of the person). This explains the first image.
Image 2 (0.714 m in front):
This arises from light refracting into the glass, reflecting off the second surface (concave), and refracting back out.
Step A: Refraction at Flat Surface (Air $\to$ Glass)
$$ v_1 = \mu u = -5\mu $$
Step B: Reflection at Curved Surface (Glass $\to$ Air interface)
The surface is concave from the outside, so from the inside (glass side), it is convex.
Focal length of this mirror $f_m = +R/2$.
Object distance for mirror $u_2 = v_1 = -5\mu$.
Mirror formula:
$$ \frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f_m} \implies \frac{1}{v_2} – \frac{1}{5\mu} = \frac{2}{R} $$
$$ \frac{1}{v_2} = \frac{2}{R} + \frac{1}{5\mu} = \frac{10\mu + R}{5\mu R} \implies v_2 = \frac{5\mu R}{10\mu + R} $$
Step C: Refraction at Flat Surface (Glass $\to$ Air)
Apparent depth formula: $v_{\text{final}} = v_2 / \mu$.
$$ v_{\text{final}} = \frac{5R}{10\mu + R} $$
Given $v_{\text{final}} = 0.714 \approx 5/7$ m:
$$ \frac{5R}{10\mu + R} = \frac{5}{7} \implies 7R = 10\mu + R \implies 6R = 10\mu \implies R = \frac{5\mu}{3} \quad \text{…(ii)} $$
4. Calculation
Equating the two expressions for $R$ from (i) and (ii): $$ 5(\mu – 1) = \frac{5\mu}{3} $$ $$ \mu – 1 = \frac{\mu}{3} $$ $$ \mu – \frac{\mu}{3} = 1 $$ $$ \frac{2\mu}{3} = 1 \implies \mu = 1.5 $$