Compound Microscope Questions

---

Very Short Answers Questions [1 Mark Questions]

Ques. Define a compound microscope.

Ans. A compound microscope is an optical instrument consisting of two suitable lenses providing a high-resolution and 2-dimensional image of the sample.

Ques. What is a simple microscope?

Ans. A simple microscope is an optical instrument to see very small objects as magnified ones. It consists of a convex lens of a small focal length. A magnifying glass is an example of a simple microscope.

Ques. The stereo microscope provides _____.

1. Two-dimensional image
2. One-dimensional image
3. Three-dimensional image
4. None of the above

Ans. The correct answer is c. Three-dimensional image

Explanation: A stereo microscope is an optical microscope designed for low magnification observation of a sample, it produces a three-dimensional visualization of the sample being examined.

Ques. What magnification does the ocular lens provide?

1. 2X
2. 3X
3. 15X
4. 50X

Ans. The correct answer is c. 15X

Explanation: Ocular lenses provide magnification of 5X, 10X, 15X, and 20X.

Ques. _____ eyepieces are seen in the binocular head.

1. Three
2. Two
3. One
4. Four

Ans. The correct answer is b. Two

Explanation: Binocular is an optical instrument used to provide a magnified stereoscopic view of distant objects. The binocular head consists of two eyepieces.

---

Short Answers Questions [2 Marks Questions]

Ques. How is the compound microscope different from a simple microscope?

Ans. A compound microscope consists of two lenses as compared to a simple microscope. A set of two lenses provides a high-resolution two-dimensional image of the sample placed in the slide. 

Ques. Define the term angular magnification of a microscope.

Ans. Angular magnification of a microscope is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the eye when both are at least a distance of distinct vision.

Ques. What are the various types of compound microscopes?

Ans. The different types of compound microscopes are

• Stereo microscopes
• Biological microscopes
• Fluorescence microscopes
• Polarizing microscopes
• Phase contrast microscopes
• Petrographic microscopes

Ques. List three uses of a compound microscope.

Ans. The three uses of compound microscopes are

• It is used to observe plant and animal cells, blood corpuscles, microorganisms like bacteria, etc.
• It is used to examine the metals and minerals in a blood sample.
• It is used in forensic labs for crime investigation.

Read More: 

---

Long Answers Questions [3 Marks Questions]

Ques. A compound microscope has an angular magnification equal to 10. If the object subtends an angle of 0.6° to the eye. What will be the angle subtended by the image to the eye?

Ans. The formula for the angular magnification of a compound microscope is given by

M_a = β/α

Where 

• β is the angle subtended by the image to the eye
• α is the angle subtended by the object to the eye
• M_P is angular magnification produced by the compound microscope

Given

• Angular magnification, M_a = 10
• α = 0.6°

On substituting the values, we get

10 = β/0.6

⇒ β = 10 x 0.6 = 6

Hence the angle subtended by the image to the eye is 6°

Ques. The tube of a microscope is of length 15 cm. Focal lengths of the eyepiece and objective are 3 cm and 2 cm respectively. Find the magnifying power of the microscope for the image formed at infinity.

Ans. Given

• Length of the microscope, L = 15 cm
• The focal length of the objective, f_o = 2 cm
• The focal length of the eyepiece, f_e = 3 cm

The formula for the magnifying power of the compound microscope when the final image is formed at infinity is given by

M_P = (L/f₀) x (D/f_e)

Where D is the least distance of distinct vision.

On substituting the values, we get

M_P = (15/2) x (25/3) = 37.5

Hence, the magnifying power of the microscope for the image formed at infinity is 37.5

Ques. A compound microscope has the objective of a focal length of 1.25 cm and an eyepiece with a focal length of 5 cm. A small object is kept at 2.5 cm from the objective. If the final image is formed at infinity, find the distance between the objective and the eyepiece.

Ans. Given

• The focal length of the objective, f_o = 1.25 cm
• The focal length of the eyepiece, f_e = 5 cm
• The distance of the object from the objective lens, u₀ = - 2.5 cm

Also, given that the final image is formed at infinity that means the object for the eyepiece is located at the focal point of the eyepiece. i.e.

u_e = f_e = 5 cm

Using the Lens formula for the objective lens

1/f₀ = -1/u₀ + 1/v₀

We get

1/v₀ = 1/f₀ + 1/u₀ = 1/1.25 - 1/2.5 = ⅓

⇒ v₀ = 3 cm

The distance between the objective and the eyepiece is, L = |v₀| + |f_e|

⇒ L = 3 + 5 = 8 cm

---

Very Long Answers Questions [5 Marks Questions]

Ques. You are given 3 lenses having powers P₁ = 6 D, P₂ = 3 D, and P₃ = 12 D. Which two of these lenses will you select to construct a compound microscope?

Ans. Let

• f₀ be the focal length of the objective lens
• f_e be the focal length of the eyepiece
• P₀ be the power of the objective lens
• P_e be the power of the eyepiece

Magnifying power (M) of a compound microscope is inversely proportional to the focal length of the objective lens and the eyepiece i.e.

M ∝ 1/f₀f_e

Also, the power of the lens is inversely proportional to the focal length i.e. P ∝ 1/f

Therefore, Magnifying power, M ∝ P₀P_e

This shows that for a compound microscope, high-power lenses are required for the objective and eyepiece but the power of the objective should be higher for a magnified and erect image.

Hence, we will choose lenses of P₃ = 12 D and P₁ = 6 D to construct a compound microscope.

In which P₃ is selected as the objective and P₁ is selected as an eyepiece.

Ques. Derive the expression of the magnifying power of a compound microscope when the final image is formed at least distance of distinct vision.

Ans. The magnifying power of a compound microscope is the product of the magnification produced by the objective lens and the eyepiece. i.e

M_P = m₀ x m_e …(i)

Where

• m₀ is the magnification of the objective lens
• m_e is the magnification of the eyepiece

Compound microscope when the final image formed at least distance of distinct vision

From the figure, we have

m₀ = A’B’/AB = v₀/-u₀ and me = A”B”/A’B’ = v/u_e

Since the final image is formed at least distance of distinct vision. Therefore v = D

⇒ m_e = D/u_e …(ii)

Now, for an eyepiece, -1/u + 1/v = 1/f_e

⇒ -1/(-u_e) + 1/(-D) = 1/f_e

⇒ D/u_e = [1 + D/f_e]

Substituting this value in equation (i), we get

m_e = [1 + D/f_e]

Again substituting the value of m0 and me in equation (i), we get

M_P = (v₀/-u₀) [1 + D/f_e]

Since the object is very close to F₀, so u₀ = f₀ 

Also, the image is formed close to the eye lens, so v₀ = L (length of the microscope)

Hence, M_P = - (L/f₀)[1 + D/f_e]

The negative sign shows that the final image is inverted.

Ques. Derive the expression of the magnifying power of a compound microscope when the final image is formed at infinity.

Ans. The magnifying power of a compound microscope is given by

M_P = m₀ x m_e …(i)

Where

• m₀ is the magnification of the objective lens
• m_e is the magnification of the eyepiece

Compound microscope when the final image is formed at infinity

The magnification produced by the eyepiece is given by

m_e = D/u_e

When image A’B’ formed by the objective is at the focus of the eyepiece (f_e) then the final image is formed at infinity. Therefore

m_e = D/f_e ….(ii)

The magnification produced by the objective lens is given by

m₀ = v₀ /-u₀

Since the object is very close to F₀ , so u₀ = f₀  

Also, the image is formed close to the eye lens, so v₀ = L (length of the microscope)

Therefore, m₀ = – L/f₀ …(iii)

Substituting equation (ii) and (iii) in equation (i), we get

M_P = (-L/f₀ ) x (D/f_e)

Hence, M_P = |v₀ /u₀ | x (D/f_e) = L/f₀ x (D/f_e)

Previous Year Questions

1. Extraction of metal from the ore cassiterite involves...[JEE Advanced 2011]
2. Commonly used vectors for human genome sequencing are...[NEET UG 2014]
3. Interfascicular cambium and cork cambium are formed due to​..
4. Pneumotaxic centre is present in​...[UP CPMT 2007]
5. Reaction of HBr with propene in the presence of peroxide gives….[NEET UG 2004]
6. Assuming the expression for the pressure exerted by the gas on the walls of the container, it can be shown that pressure is...[MHT CET 2016]
7. Which among the following is the strongest acid?...[TS EAMCET 2017]
8. Isopropyl alcohol on oxidation forms​..
9. A vector is not changed if​..
10. Which of the following arrangements does not represent the correct order of the property stated against it?...[JEE Main 2013]

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates

---

Check-Out: