Refer the figure below. \( \mu_1 \) and \( \mu_2 \) are refractive indices of air and lens material respectively. The height of image will be _____ cm.
Step 1: Understanding the Concept:
The problem involves refraction at a single spherical surface.
First, we find the image position \(v\) using the spherical refraction formula.
Then, we calculate the lateral magnification to find the height of the image. Step 2: Key Formula or Approach:
1. Refraction formula: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]
2. Lateral magnification formula: \[ m = \frac{h_i}{h_o} = \frac{\mu_1 v}{\mu_2 u} \] Step 3: Detailed Explanation:
From the provided figure and data:
Refractive index of object medium (air) \(\mu_1 = 1\).
Refractive index of image medium \(\mu_2 = 1.54\).
Object distance \(u = -40 \text{ cm}\) (point \(O\) is 40 cm from vertex \(P\)).
Radius of curvature \(R = -20 \text{ cm}\) (concave surface with center of curvature at \(C\)).
Height of object \(h_o = 2 \text{ cm}\).
Applying the refraction formula:
\[ \frac{1.54}{v} - \frac{1}{-40} = \frac{1.54 - 1}{-20} \]
\[ \frac{1.54}{v} + 0.025 = \frac{0.54}{-20} \]
\[ \frac{1.54}{v} + 0.025 = -0.027 \]
\[ \frac{1.54}{v} = -0.027 - 0.025 = -0.052 \]
\[ v = \frac{1.54}{-0.052} \approx -29.61 \text{ cm} \].
Now, calculate the magnification:
\[ m = \frac{\mu_1 v}{\mu_2 u} = \frac{1 \times (-29.61)}{1.54 \times (-40)} \]
\[ m = \frac{29.61}{61.6} \approx 0.48 \]
The height of the image \(h_i\) is:
\[ h_i = m \times h_o = 0.48 \times 2 = 0.96 \text{ cm} \].
Rounding off to the nearest integer as per the options provided, we get 1 cm. Step 4: Final Answer:
The height of the image is 1 cm.
