Optical Instruments — Microscope, telescope
Ray Optics
12
JEE Qs
8%
Hard
75
min
Master the ray diagrams and understand the two extreme cases (image at infinity and at D) for each instrument, as these form the basis for all derivations and problem-solving.
🧮 Key Formulas
✅ Key Points for JEE
- 1Angular magnification is the primary figure of merit for optical instruments, representing how much larger the image appears to the eye compared to the object viewed directly.
- 2Understand and be able to draw ray diagrams for both simple and compound microscopes, and refracting telescopes, distinguishing between 'normal adjustment' (image at infinity) and 'final image at least distance of distinct vision (D)'.
- 3The objective lens forms a real, inverted, magnified intermediate image which then acts as the object for the eyepiece lens.
- 4Reflecting telescopes offer advantages over refracting telescopes: no chromatic aberration, reduced spherical aberration (parabolic mirrors), and easier mechanical support for large apertures.
⚠️ Common Mistakes
- ✕Confusing linear magnification with angular magnification, especially in problems where the final image is viewed by the eye.
- ✕Incorrect application of sign conventions in the lens formula, leading to errors in calculating image distances or focal lengths.
- ✕Mixing up formulas for magnification and length for different adjustment conditions (normal vs. final image at D) or for different instruments.
- ✕Forgetting the role of the intermediate image formed by the objective lens in compound instruments.
📝 Practice Questions
See allQ46.The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature R = 2 m . Another car approaches him from behind with a uniform speed of 90 km/hr. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the car in the side view mirror is ' a '. The value of 100 a is _______ m/s2 .
Q27.A spherical surface of radius of curvature R, separates air from glass (refractive index = 1.5 ). The centre of curvature is in the glass medium. A point object ' O ' placed in air on the optic axis of the surface, so that its real image is formed at ' I ' inside glass. The line OI intersects the spherical surface at P and PO = PI. The distance PO equals to (1) 5 R (2) 3 R (3) 1.5 R (4) 2 R
Q32.Given is a thin convex lens of glass (refractive index μ ) and each side having radius of curvature R. One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that 2025 (22 Jan Shift 1) JEE Main Previous Year Paper the image gets formed on the object itself ? (1) R/μ (2) R/(2μ −3) (3) μR (4) R/(2μ −1)
Q42.In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to |R1| and |R2|, i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is (1) 1 − + |R2| |R2| 6 ( |R1|1 1 ) (2) −16 ( |R1|1 1 ) (3) 1 + − |R2| |R2| 6 ( |R1|1 1 ) (4) −16 ( |R1|1 1 )
Q29.A symmetric thin biconvex lens is cut into four equal parts by two planes AB and CD as shown in figure. If the power of original lens is 4 D then the power of a part of the divided lens is (1) D (2) 8D (3) 2D (4) 4D
Q29.Given a thin convex lens (refractive index μ2 ), kept in a liquid (refractive index μ1, μ1 < μ2 ) having radii of curvatures |R1| and |R2|. Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place? (1) μ1|R1|⋅|R2| (2) μ1|R1|⋅|R2| μ2(|R1|+|R2|)−μ1|R2| μ2(|R1|+|R2|)−μ1|R1| (3) (μ2+μ1)|R1| (4) μ1|R1|⋅|R2| (μ2−μ1) μ2(2|R1|+|R2|)−μ1√|R1|⋅|R2|
NCERT Chapters
- Class 12 Physics Ch 9: Ray Optics and Optical Instruments