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Ray Optics and Optical Instruments - Microscopes and Telescopes

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Simple Microscope: A single convex lens of short focal length. It produces an erect, virtual, and magnified image when the object is placed within its focal length.

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Least Distance of Distinct Vision (DD): For a normal human eye, the value of DD is 25 cm25\text{ cm}.

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Compound Microscope: Consists of two convex lenses. The lens near the object is the objective (short focal length fof_o and small aperture), and the lens near the eye is the eyepiece (moderate focal length fef_e and larger aperture).

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Magnification of Compound Microscope: The total magnification mm is the product of the linear magnification of the objective (mom_o) and the angular magnification of the eyepiece (mem_e).

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Astronomical (Refracting) Telescope: Used to view distant objects. The objective has a large focal length fof_o and large aperture to collect more light, while the eyepiece has a small focal length fef_e.

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Normal Adjustment (Telescope): The state where the final image is formed at infinity. In this case, the distance between lenses is L=fo+feL = f_o + f_e.

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Reflecting Telescope (Cassegrain): Uses a concave parabolic primary mirror and a secondary mirror. Advantages include no chromatic aberration and reduced spherical aberration compared to refracting telescopes.

📐Formulae

m=1+Df (Simple Microscope: Image at D)m = 1 + \frac{D}{f} \text{ (Simple Microscope: Image at } D\text{)}

m=Df (Simple Microscope: Image at infinity)m = \frac{D}{f} \text{ (Simple Microscope: Image at infinity)}

m=mo×me=(vo−uo)(Due)m = m_o \times m_e = \left( \frac{v_o}{-u_o} \right) \left( \frac{D}{u_e} \right)

m≈Lfo(1+Dfe) (Compound Microscope: Image at D)m \approx \frac{L}{f_o} \left( 1 + \frac{D}{f_e} \right) \text{ (Compound Microscope: Image at } D\text{)}

m≈Lfo⋅Dfe (Compound Microscope: Image at infinity)m \approx \frac{L}{f_o} \cdot \frac{D}{f_e} \text{ (Compound Microscope: Image at infinity)}

m=fofe (Telescope: Normal Adjustment)m = \frac{f_o}{f_e} \text{ (Telescope: Normal Adjustment)}

m=fofe(1+feD) (Telescope: Image at D)m = \frac{f_o}{f_e} \left( 1 + \frac{f_e}{D} \right) \text{ (Telescope: Image at } D\text{)}

L=fo+fe (Length of Telescope in Normal Adjustment)L = f_o + f_e \text{ (Length of Telescope in Normal Adjustment)}

💡Examples

Problem 1:

A compound microscope has an objective lens of focal length 2.0 cm2.0\text{ cm} and an eyepiece of focal length 6.25 cm6.25\text{ cm} separated by a distance of 15 cm15\text{ cm}. How far from the objective should an object be placed in order to obtain the final image at the least distance of distinct vision (25 cm25\text{ cm})?

Solution:

  1. For the eyepiece: ve=−25 cmv_e = -25\text{ cm}, fe=6.25 cmf_e = 6.25\text{ cm}. Using lens formula 1ve−1ue=1fe\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}, we get 1ue=1−25−16.25=−525⇒ue=−5 cm\frac{1}{u_e} = \frac{1}{-25} - \frac{1}{6.25} = -\frac{5}{25} \Rightarrow u_e = -5\text{ cm}.
  2. The distance between lenses L=∣vo∣+∣ue∣=15 cmL = |v_o| + |u_e| = 15\text{ cm}. Thus, ∣vo∣=15−5=10 cm|v_o| = 15 - 5 = 10\text{ cm}. Since the objective forms a real image, vo=+10 cmv_o = +10\text{ cm}.
  3. For the objective: vo=10 cmv_o = 10\text{ cm}, fo=2.0 cmf_o = 2.0\text{ cm}. Using 1vo−1uo=1fo\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}, we get 1uo=110−12=−410⇒uo=−2.5 cm\frac{1}{u_o} = \frac{1}{10} - \frac{1}{2} = -\frac{4}{10} \Rightarrow u_o = -2.5\text{ cm}.

Explanation:

The object must be placed 2.5 cm2.5\text{ cm} in front of the objective lens. We first find the object distance for the eyepiece using the image position (DD), then use the tube length to find the image distance for the objective.

Problem 2:

An astronomical telescope has an angular magnification of 1010 for distant objects. The separation between the objective and the eyepiece is 44 cm44\text{ cm} in normal adjustment. Calculate fof_o and fef_e.

Solution:

  1. For normal adjustment, m=fofe=10⇒fo=10fem = \frac{f_o}{f_e} = 10 \Rightarrow f_o = 10 f_e.
  2. The length of the telescope L=fo+fe=44 cmL = f_o + f_e = 44\text{ cm}.
  3. Substitute fof_o: 10fe+fe=44⇒11fe=44⇒fe=4 cm10 f_e + f_e = 44 \Rightarrow 11 f_e = 44 \Rightarrow f_e = 4\text{ cm}.
  4. Then, fo=10×4=40 cmf_o = 10 \times 4 = 40\text{ cm}.

Explanation:

In normal adjustment, the focal point of the objective coincides with the focal point of the eyepiece, making the total length the sum of their focal lengths.

Microscopes and Telescopes - Revision Notes & Key Formulas | CBSE Class 12 Physics