
A luminous object and a screen are at a fixed distance D apart. A converging lens of focal length f is placed between the object and screen. A real image of the object is formed on the screen for two lens positions if they are separated by a distance d equal to
A. $\sqrt{D(D+4f)} \\ $
B. $\sqrt{D(D-4f)} \\ $
C. $\sqrt{2D(D-4f)} \\ $
D. $\sqrt{{{D}^{2}}+4f}$
Answer
164.1k+ views
Hint: In this question, we have to assume the object distance as $x$ and the image distance will become $D-x$ as the lens of focal length $f$ is placed in between the object and a screen. After applying the lens formula we get the required answer.
Formula used:
Lens equation is used.
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
Where f is the focal length of the lens, u is the object distance and v is the image distance.
Complete step by step solution:
It is given that the luminous object and screen are placed D distance apart. A converging lens of focal length f is placed between object and screen. Let the object distance be x. Then, the image distance is D-x. Let d be the separation of distance of two lens positions for which real image is formed.
Now by applying lens equation we get:
$\dfrac{1}{x}+\dfrac{1}{D-x}=\dfrac{1}{f}$
On doing further simplification, we get distance D as
${{x}^{2}}-Dx+Df=0$
On solving this quadratic equation, we get:
$x=\dfrac{D-\sqrt{D(D-4f)}}{2} \\ $
$\Rightarrow x'=\dfrac{D+\sqrt{D(D-4f)}}{2}$
Therefore, the distance between the two object positions is
$\therefore d=x'-x=\sqrt{D(D-4f)}$
That is, a real image of the object is formed on the screen for two lens positions if they are separated by a distance d equal to $\sqrt{D(D-4f)}$.
Therefore, the correct answer is option B.
Notes: We have to apply sign conversion to the lens equation. Remember we have to find the distance of separation not the distance of formation of real image. We also have to note that converging lenses are used. Note that mirror formula, lens formula and lens maker’s formula are different.
Formula used:
Lens equation is used.
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
Where f is the focal length of the lens, u is the object distance and v is the image distance.
Complete step by step solution:
It is given that the luminous object and screen are placed D distance apart. A converging lens of focal length f is placed between object and screen. Let the object distance be x. Then, the image distance is D-x. Let d be the separation of distance of two lens positions for which real image is formed.
Now by applying lens equation we get:
$\dfrac{1}{x}+\dfrac{1}{D-x}=\dfrac{1}{f}$
On doing further simplification, we get distance D as
${{x}^{2}}-Dx+Df=0$
On solving this quadratic equation, we get:
$x=\dfrac{D-\sqrt{D(D-4f)}}{2} \\ $
$\Rightarrow x'=\dfrac{D+\sqrt{D(D-4f)}}{2}$
Therefore, the distance between the two object positions is
$\therefore d=x'-x=\sqrt{D(D-4f)}$
That is, a real image of the object is formed on the screen for two lens positions if they are separated by a distance d equal to $\sqrt{D(D-4f)}$.
Therefore, the correct answer is option B.
Notes: We have to apply sign conversion to the lens equation. Remember we have to find the distance of separation not the distance of formation of real image. We also have to note that converging lenses are used. Note that mirror formula, lens formula and lens maker’s formula are different.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

Instantaneous Velocity - Formula based Examples for JEE
