SATHEE: Optics Question 817

Optics Question 817

Question: A luminous object and a screen are at a fixed distance D apart. A converging lens of focal length l is placed between the object and screen. A real image of the object in formed on the screen for two lens positions of they are separated by a distance d equal to

Options:

A) $\sqrt{D (D + 4 f)}$

B) $\sqrt{D (D - 4 f)}$

C) $\sqrt{2 D (D - 4 f)}$

D) $\sqrt{D^{2} + 4 f}$

Show Answer

Answer:

Correct Answer: B

Solution:

[b] Let the object distance be x. Then, the image distance is $D - x$ .

From lens equation, $\frac{1}{x} + \frac{1}{D - x} = \frac{1}{f}$

On algebraic rearrangement, we get $x^{2} - D x + D f = 0$

On solving for x, we get $x_{1} = \frac{D - \sqrt{D (D - 4 f)}}{2} x_{2} = \frac{D + \sqrt{D (D - 4 f)}}{2}$

The distance between the two object positions is $d = x_{2} - x_{1} = \sqrt{D (D - 4 f)}$

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Optics Question 817

Question: A luminous object and a screen are at a fixed distance D apart. A converging lens of focal length l is placed between the object and screen. A real image of the object in formed on the screen for two lens positions of they are separated by a distance d equal to

Options:

Answer:

Solution:

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