Start Learning Free

Before You Begin

Question #43228

Difficulty: 8/10 | 🟢 Free

The Question

A compound microscope is constructed using an objective lens of focal length fo=2.0 cmf_o = 2.0 \text{ cm} and an eyepiece of focal length fe=6.0 cmf_e = 6.0 \text{ cm}. An object is placed at a distance of 2.5 cm2.5 \text{ cm} in front of the objective. To make the instrument compact, a right-angled glass prism of refractive index n=1.5n = 1.5 is inserted between the lenses to fold the light path via Total Internal Reflection. The physical length of the light path inside the glass prism is t=6.0 cmt = 6.0 \text{ cm}. If the final virtual image is formed at a distance of D=30.0 cmD = 30.0 \text{ cm} from the eyepiece, calculate the total physical path length LL of the central light ray between the objective and the eyepiece. (Assume paraxial rays and that the unfolded prism acts equivalently to a glass slab of thickness tt).

Options

  • A.

    17.0 cm17.0 \text{ cm}

  • B.

    12.0 cm12.0 \text{ cm}

  • C.

    15.0 cm15.0 \text{ cm}

  • D.

    19.5 cm19.5 \text{ cm}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Microscope Tube Length|Total Internal Reflection (TIR)|Sign Convention Trap

Why This Question Is Tricky

Common Mistakes

This problem tests two critical JEE Advanced concepts: the effect of a dielectric medium on image position and the Gaussian sign convention for virtual images.

Students often fall into the 'Optical Path Length' trap by incorrectly using Leff=L−t+ntL_{eff} = L - t + nt (adding optical path for phase) instead of the apparent depth shift Δx=t(1−1/n)\Delta x = t(1 - 1/n), leading to the incorrect answer of 12.0 cm12.0 \text{ cm}.

Additionally, the 'Sign Convention Trap' catches students who substitute ve=+30.0 cmv_e = +30.0 \text{ cm} for the virtual image, yielding an incorrect eyepiece object distance of 7.5 cm7.5 \text{ cm} and a final answer of 19.5 cm19.5 \text{ cm}.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterRay Optics and Optical Instruments - Class 12
ConceptMicroscope Tube Length
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree