Q25.The value of numerical aperture of the objective lens of a microscope is 1.25 . If light of wavelength 5000 Å is used, the minimum separation between two points, to be seen as distinct, will be: (1) 0.48 μm (2) 0.24 μm (3) 0.38 μm (4) 0.12 μm
What This Question Tests
This question is a direct application of the formula for the minimum separation (resolving power) of a microscope, given its numerical aperture and the wavelength of light.
Concepts Tested
Formulas Used
d_min = λ / (2 * NA)
📚 NCERT Sections This Tests
9.26 — Assume Microscope In Normal Use I.E., Image At 25 Cm. Angular
Physics Class 12 · Chapter 9
9.26 Assume microscope in normal use i.e., image at 25 cm. Angular magnification of the eye-piece 25 = 1 6 5 Magnification of the objective 30 = 5 6 1 1 1 − = 5u O u O 1.25 which gives uO= –1.5 cm; v0= 7.5 cm. |ue| (25/6) cm = 4.17 cm. The separation between the objective and the eye-piece should be (7.5 + 4.17) cm = 11.67 cm. Further the object should be placed 1.5 cm from the objective to obtain the desired magnification.
10.4 — In A Young’S Double-Slit Experiment, The Slits Are Separated By
Physics Class 12 · Chapter 10
10.4 In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
9.27 — (A) M = ( Fo/Fe) = 28
Physics Class 12 · Chapter 9
9.27 (a) m = ( fO/fe) = 28 f O f O (b) m = 1 + = 33.6 f e 25 349 Reprint 2025-26 Physics 9.28 (a) fO + fe = 145 cm (b) Angle subtended by the tower = (100/3000) = (1/30) rad. Angle subtended by the image produced by the objective h h = = f O 140 Equating the two, h = 4.7 cm. (c) Magnification (magnitude) of the eye-piece = 6. Height of the final image (magnitude) = 28 cm. 9.29 The image formed by the larger (concave) mirror acts as virtual object for the smaller (convex) mirror. Parallel rays coming from the object at infinity will focus at a distance of 110 mm from the larger mirror. The distance of virtual object for the smaller mirror = (110 –20) = 90 mm. The focal length of smaller mirror is 70 mm. Using the mirror formula, image is formed at 315 mm from the smaller mirror. 9.30 The reflected rays get deflected by twice the angle of rotation of the mirror. Therefore, d/1.5 = tan 7°. Hence d = 18.4 cm. 9.31 n = 1.33 CHAPTER 10 10.1 (a) Reflected light: (wavelength, frequency, speed same as incident light) l = 589 nm, n = 5.09 ´ 1014 Hz, c = 3.00 ´ 108 m s–1 (b) Refracted light: (frequency same as the incident frequency) n = 5.09 ´ 1014Hz v = (c/n) = 2.26 × 108 m s–1, l = (v/n) = 444 nm 10.2 (a) Spherical (b) Plane (c) Plane (a small area on the surface of a large sphere is nearly planar). 10.3 (a) 2.0 × 108 m s–1 (b) No. The refractive index, and hence the speed of light in a medium, depends on wavelength. [When no particular wavelength or colour of light is specified, we may take the given refractive index to refer to yellow colour.] Now we know violet colour deviates more than red in a glass prism, i.e. nv > nr. Therefore, the violet component of white light travels slower than the red component. 1.2 10 – 2 0.28 10 – 3 10.4 m = 600 nm 4 14. 10.5 K/4 10.6 (a) 1.17 mm (b) 1.56 mm 10.7 0.15° 350 10.8 tan–1(1.5) ~ 56.3o Reprint 2025-26 Answers
📋 Question Details
- Chapter
- Wave Optics
- Topic
- Resolution of Microscope
- Year
- 2019
- Shift
- 12 Apr Shift 1
- Q Number
- Q25
- Type
- MCQ
- NCERT Ref
- Class 12 Physics Ch 10: Wave Optics
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