Angle made by image of inclined rod with principal axis of convex lens

To find the angle $\alpha$ that the image of the inclined rod makes with the principal axis, we analyze the coordinates of the two endpoints of the rod in object space and determine their corresponding image positions.

1. Coordinates of the Object Endpoints

Let the origin $(0,0)$ be at the optical center of the convex lens, with the principal axis along the x-axis and light traveling from left to right (positive x-direction).

• Endpoint $A$ lies on the principal axis at a distance of $\frac{40}{3}\text{ cm}$ in front of the lens: $u_A = -\frac{40}{3}\text{ cm}, \quad y_A = 0\text{ cm}$

• Endpoint $B$ is the upper end of the rod of length $L = 2\text{ cm}$ inclined at an angle of $\frac{2\pi}{3}\text{ rad} = 120^\circ$ relative to the positive x-axis: $\Delta x = L \cos\left(\frac{2\pi}{3}\right) = 2 \times \left(-\frac{1}{2}\right) = -1\text{ cm}$ $\Delta y = L \sin\left(\frac{2\pi}{3}\right) = 2 \times \left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}\text{ cm}$

Thus, the position of endpoint $B$ is: $u_B = u_A + \Delta x = -\frac{40}{3} - 1 = -\frac{43}{3}\text{ cm}$ $y_B = \sqrt{3}\text{ cm}$

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2. Image of Endpoint $A$ ($A'$)

Using the thin lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with focal length $f = +10\text{ cm}$:

$\frac{1}{v_A} - \frac{1}{-40/3} = \frac{1}{10}$ $\frac{1}{v_A} = \frac{1}{10} - \frac{3}{40} = \frac{1}{40} \implies v_A = 40\text{ cm}$

So, the image point $A'$ lies on the principal axis at: $x_{A'} = 40\text{ cm}, \quad y_{A'} = 0\text{ cm}$

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3. Image of Endpoint $B$ ($B'$)

Applying the thin lens formula for endpoint $B$:

$\frac{1}{v_B} - \frac{1}{-43/3} = \frac{1}{10}$ $\frac{1}{v_B} = \frac{1}{10} - \frac{3}{43} = \frac{13}{430} \implies v_B = \frac{430}{13}\text{ cm}$

The transverse magnification for point $B$ is: $m_B = \frac{v_B}{u_B} = \frac{430/13}{-43/3} = -\frac{30}{13}$

The y-coordinate of the image $B'$ is: $y_{B'} = m_B \cdot y_B = -\frac{30}{13} \times \sqrt{3} = -\frac{30\sqrt{3}}{13}\text{ cm}$

So, the position of image point $B'$ is: $x_{B'} = \frac{430}{13}\text{ cm}, \quad y_{B'} = -\frac{30\sqrt{3}}{13}\text{ cm}$

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4. Calculating Angle $\alpha$

The angle $\alpha$ that the image line $A'B'$ makes with the principal axis is given by:

$\tan \alpha = \frac{|y_{A'} - y_{B'}|}{|x_{A'} - x_{B'}|}$

Substituting the values: $|x_{A'} - x_{B'}| = 40 - \frac{430}{13} = \frac{520 - 430}{13} = \frac{90}{13}\text{ cm}$ $|y_{A'} - y_{B'}| = \frac{30\sqrt{3}}{13}\text{ cm}$

$\tan \alpha = \frac{\frac{30\sqrt{3}}{13}}{\frac{90}{13}} = \frac{30\sqrt{3}}{90} = \frac{1}{\sqrt{3}}$

Taking the inverse tangent: $\alpha = \frac{\pi}{6}\text{ rad}$

Comparing this with $\alpha = \frac{\pi}{n}\text{ rad}$, we get: $n = 6$