Question

A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is:

• A. \( \frac{-\alpha}{2} \)
• B. \( -45^\circ \)
• C. \( +45^\circ \)
• D. \( -\alpha \)

Hint:

When the object is slanted and small compared to the location, use magnification and lens formulas to find the angle made by the image.

Answer 1

The Correct Option is B

The location of the image of A can be found using the lens formula: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] where \( f = 20 \, \text{cm} \), \( u = -30 \, \text{cm} \), and \( v = 60 \, \text{cm} \). Using the magnification formula: \[ m = \frac{v}{u} = \frac{60}{-30} = -2 \] Since the object size is small with respect to the location, we can calculate the small change \( dv \) in the image: \[ dv = m^2 du = 4 \times 1 = 4 \, \text{cm} \] This gives us the size of the image at \( P \) as \( h_i = m h_o = 2 \times 2 = 4 \, \text{cm} \). 
The angle made by the image with the principal axis is \( -45^\circ \), which corresponds to the correct answer.

Answer 2

The problem asks for the angle made by the image of a slanted object AB with the principal axis. The object is placed in front of a convex lens. We can solve this by finding the coordinates of the images of the endpoints A and B of the object and then determining the slope of the line connecting these image points.

Concept Used:

The solution relies on the thin lens formula and the lateral magnification formula.

1. Thin Lens Formula: This formula relates the object distance (\(u\)), the image distance (\(v\)), and the focal length (\(f\)) of the lens. \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] We use the sign convention where the optical center is the origin, and light travels from left to right. Distances measured against the direction of light are negative.
2. Lateral Magnification (m): This gives the ratio of the height of the image (\(h_i\)) to the height of the object (\(h_o\)). \[ m = \frac{h_i}{h_o} = \frac{v}{u} \]
3. Slope of a line: The angle \( \beta \) that a line segment (the image) makes with the principal axis (x-axis) can be found from its slope, which is given by \( \tan(\beta) = \frac{\Delta y}{\Delta x} \), where \( \Delta x \) and \( \Delta y \) are the differences in the x and y coordinates of the endpoints of the segment.

 

Step-by-Step Solution:

Step 1: Determine the coordinates of the endpoints of the object (A and B).

Let the optical center of the convex lens be the origin (0, 0). The principal axis is the x-axis. According to the sign convention and the diagram:

• Focal length of the convex lens, \( f = +20 \, \text{cm} \).
• Point A is on the principal axis, at a distance of 30 cm from the lens. So, its coordinates are \( A = (-30, 0) \).
• Point B is 1 cm to the right of A and 2 cm above A. So, its x-coordinate is \( -30 + 1 = -29 \, \text{cm} \) and its y-coordinate is \( 2 \, \text{cm} \). The coordinates of B are \( B = (-29, 2) \).

Step 2: Find the coordinates of the image of point A (let's call it A').

For point A, the object distance is \( u_A = -30 \, \text{cm} \). Using the lens formula to find the image distance \( v_A \):

\[ \frac{1}{v_A} - \frac{1}{-30} = \frac{1}{20} \] \[ \frac{1}{v_A} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60} \] \[ v_A = +60 \, \text{cm} \]

Since point A is on the principal axis, its image A' will also be on the principal axis. The coordinates of A' are \( (60, 0) \).

Step 3: Find the coordinates of the image of point B (let's call it B').

For point B, the object distance is \( u_B = -29 \, \text{cm} \) and the object height is \( h_{oB} = 2 \, \text{cm} \). First, we find the image distance \( v_B \) using the lens formula:

\[ \frac{1}{v_B} - \frac{1}{-29} = \frac{1}{20} \] \[ \frac{1}{v_B} = \frac{1}{20} - \frac{1}{29} = \frac{29 - 20}{20 \times 29} = \frac{9}{580} \] \[ v_B = \frac{580}{9} \, \text{cm} \]

Now, we find the height of the image of B (\( h_{iB} \)) using the lateral magnification formula:

\[ m_B = \frac{v_B}{u_B} = \frac{580/9}{-29} = -\frac{580}{9 \times 29} = -\frac{20}{9} \] \[ h_{iB} = m_B \times h_{oB} = \left(-\frac{20}{9}\right) \times 2 = -\frac{40}{9} \, \text{cm} \]

The coordinates of B' are \( \left(\frac{580}{9}, -\frac{40}{9}\right) \).

Step 4: Calculate the slope of the image A'B' to find the angle.

The image is the line segment connecting A'(60, 0) and B'\( \left(\frac{580}{9}, -\frac{40}{9}\right) \). Let \( \beta \) be the angle the image makes with the principal axis. The tangent of this angle is the slope of the line A'B'.

\[ \tan(\beta) = \frac{\Delta y}{\Delta x} = \frac{y_{B'} - y_{A'}}{x_{B'} - x_{A'}} \] \[ \tan(\beta) = \frac{-\frac{40}{9} - 0}{\frac{580}{9} - 60} = \frac{-\frac{40}{9}}{\frac{580 - 540}{9}} = \frac{-\frac{40}{9}}{\frac{40}{9}} = -1 \]

Final Computation & Result:

The tangent of the angle made by the image with the principal axis is -1.

\[ \tan(\beta) = -1 \]

This means the angle of inclination with the positive x-axis is \( 135^\circ \). The question asks for the angle made by the image with the principal axis, which is typically the acute angle. The magnitude of the angle is:

\[ \beta = \tan^{-1}(|-1|) =- 45^\circ \]

The angle made by the image with the principal axis is - 45°.