A slanted object AB is placed on one side of convex lens as shown… — Solution | JEE Physics

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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 convex lens on the principal axis with object AB on the left; A lies 30 cm from the lens on the axis, B is 1 cm right and 2 cm above A, angle alpha at A, and focal length f = 20 cm marked on the right.

A
$\frac{-\alpha}{2}$
B
$-45^\circ$
C
$+45^\circ$
D
$-\alpha$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A convex lens has $f = 20 \, \text{cm}$ . Point $A$ of the object is at $u_A = -30 \, \text{cm}$ on the principal axis, and point $B$ is $1 \, \text{cm}$ to the right and $2 \, \text{cm}$ above $A$ , so $B = (-29, 2)$ .

Find: The angle made by the image $A'B'$ with the principal axis.

Using the thin lens formula for point $A$ :

\frac{1}{v_A} - \frac{1}{u_A} = \frac{1}{f}

\frac{1}{v_A} - \frac{1}{-30} = \frac{1}{20}

\frac{1}{v_A} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60}

v_A = 60 \, \text{cm}

So, the image of $A$ is $A' = (60, 0)$ .

Now for point $B$ , the object distance is $u_B = -29 \, \text{cm}$ . Again using the lens formula:

\frac{1}{v_B} - \frac{1}{u_B} = \frac{1}{f}

\frac{1}{v_B} - \frac{1}{-29} = \frac{1}{20}

\frac{1}{v_B} = \frac{1}{20} - \frac{1}{29} = \frac{9}{580}

v_B = \frac{580}{9} \, \text{cm}

Slope of the image

Using magnification for point $B$ :

m_B = \frac{v_B}{u_B} = \frac{580/9}{-29} = -\frac{20}{9}

If the object height of point $B$ above the axis is $h_{oB} = 2 \, \text{cm}$ , then

h_{iB} = m_B h_{oB} = \left(-\frac{20}{9}\right) \times 2 = -\frac{40}{9} \, \text{cm}

Hence,

B' = \left(\frac{580}{9}, -\frac{40}{9}\right)

Small-shift method

Since the object is small compared to its location, use the local magnification near $A$ . From the solution working,

m = \frac{v}{u} = \frac{60}{-30} = -2

and for a small horizontal shift $du = 1 \, \text{cm}$ ,

dv = m^2 du = 4 \times 1 = 4 \, \text{cm}

Also, the vertical size transforms as

h_i = |m| h_o = 2 \times 2 = 4 \, \text{cm}

So the image has horizontal change $4 \, \text{cm}$ and vertical change $-4 \, \text{cm}$ , giving slope

\tan \beta = \frac{-4}{4} = -1

Therefore, the image makes an angle of $-45^\circ$ with the principal axis. The correct option is B.

Common mistakes

Using the same image distance for both endpoints $A$ and $B$ is incorrect because the object is slanted, so their object distances are different. Apply the lens formula separately to each endpoint or use the small-shift relation carefully.
Ignoring the sign convention leads to the wrong orientation of the image. Here $u < 0$ for the object on the left, and the image height comes out negative, which shows inversion. Keep signs throughout the lens and magnification formulas.
Computing only the magnified height and forgetting the horizontal shift gives the wrong angle. The angle of the image depends on both $\Delta x$ and $\Delta y$ , so use the slope $\tan \beta = \frac{\Delta y}{\Delta x}$ .

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Answer

Step-by-step solution

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