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Optics - Optical Instruments (Microscopes and Telescopes)

Grade 12ICSEPhysics

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

🔑Concepts

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The Visual Angle is the angle subtended by an object at the eye. Optical instruments increase this angle to make the object appear larger.

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A Simple Microscope consists of a single convex lens of short focal length. The object is placed within the focal point to produce an erect, virtual, and magnified image.

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A Compound Microscope uses two lenses: the Objective (short focal length fof_o and small aperture) and the Eyepiece (larger focal length fef_e and larger aperture). The intermediate image is real, inverted, and magnified.

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The Astronomical Telescope is used to view distant objects. The objective lens has a large focal length fof_o and a large aperture to collect more light, while the eyepiece has a small focal length fef_e.

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Normal Adjustment refers to the state where the final image is formed at infinity (v=∞v = \infty). In this state, the eye is most relaxed.

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Least Distance of Distinct Vision (DD) is the minimum distance at which the eye can see objects clearly without strain, taken as D=25 cmD = 25\text{ cm} for a normal eye.

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Reflecting Telescopes (like the Cassegrain telescope) use a concave parabolic mirror as the objective to eliminate chromatic aberration and reduce spherical aberration.

📐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 ∞)M = \frac{D}{f} \text{ (Simple Microscope, image at } \infty\text{)}

M=mo×me=(−vouo)(1+Dfe) (Compound Microscope, image at D)M = m_o \times m_e = \left( -\frac{v_o}{u_o} \right) \left( 1 + \frac{D}{f_e} \right) \text{ (Compound Microscope, image at } D\text{)}

M≈−Lfo(1+Dfe) (Where L is the length of the tube)M \approx -\frac{L}{f_o} \left( 1 + \frac{D}{f_e} \right) \text{ (Where } L \text{ is the length of the tube)}

M=−fofe (Astronomical Telescope, normal adjustment)M = -\frac{f_o}{f_e} \text{ (Astronomical Telescope, normal adjustment)}

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

M=−fofe(1+feD) (Astronomical Telescope, image at D)M = -\frac{f_o}{f_e} \left( 1 + \frac{f_e}{D} \right) \text{ (Astronomical Telescope, image at } D\text{)}

Resolving Power of Microscope=2μsin⁡θλ\text{Resolving Power of Microscope} = \frac{2 \mu \sin \theta}{\lambda}

Resolving Power of Telescope=D1.22λ\text{Resolving Power of Telescope} = \frac{D}{1.22 \lambda}

💡Examples

Problem 1:

A compound microscope has an objective of focal length 1.0 cm1.0\text{ cm} and an eyepiece of focal length 2.5 cm2.5\text{ cm}. An object is placed at a distance of 1.1 cm1.1\text{ cm} from the objective. Calculate the magnifying power of the microscope if the final image is formed at D=25 cmD = 25\text{ cm}.

Solution:

Given: fo=1.0 cmf_o = 1.0\text{ cm}, fe=2.5 cmf_e = 2.5\text{ cm}, uo=−1.1 cmu_o = -1.1\text{ cm}. Using lens formula for objective: 1vo−1uo=1fo⇒1vo=11.0+1−1.1=1−0.909=0.091\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o} \Rightarrow \frac{1}{v_o} = \frac{1}{1.0} + \frac{1}{-1.1} = 1 - 0.909 = 0.091 vo=11 cmv_o = 11\text{ cm}. Now, magnification M=−vouo(1+Dfe)M = -\frac{v_o}{u_o} \left( 1 + \frac{D}{f_e} \right) M=−11−1.1(1+252.5)=10×(1+10)=110M = -\frac{11}{-1.1} \left( 1 + \frac{25}{2.5} \right) = 10 \times (1 + 10) = 110.

Explanation:

First, find the image distance for the objective lens using the lens formula. Then, substitute values into the magnifying power formula for the compound microscope when the image is at the least distance of distinct vision.

Problem 2:

An astronomical telescope has an angular magnification of magnitude 55 for distant objects. The separation between the objective and the eyepiece is 36 cm36\text{ cm} and the final image is formed at infinity. Find the focal lengths of the lenses.

Solution:

In normal adjustment (image at infinity): M=fofe=5⇒fo=5feM = \frac{f_o}{f_e} = 5 \Rightarrow f_o = 5 f_e L=fo+fe=36 cmL = f_o + f_e = 36\text{ cm} Substitute fof_o: 5fe+fe=36⇒6fe=36⇒fe=6 cm5f_e + f_e = 36 \Rightarrow 6f_e = 36 \Rightarrow f_e = 6\text{ cm}. Then, fo=5×6=30 cmf_o = 5 \times 6 = 30\text{ cm}.

Explanation:

In normal adjustment for a telescope, the magnification is the ratio of focal lengths and the length of the tube is the sum of the focal lengths. Solving these two simultaneous equations yields the individual focal lengths.

Optical Instruments (Microscopes and Telescopes) Revision - Class 12 Physics ICSE