Question

The intermediate image formed in a compound microscope is:

• A. real, inverted and magnified
• B. real, erect and magnified
• C. virtual, erect and magnified
• D. virtual, inverted and magnified
• A. 3.45 cm
• B. 5 cm
• C. 1.29 cm
• D. 2.59 cm
• A. 7.47 cm
• B. 8.47 cm
• C. 9.47 cm
• D. 10.47 cm
• A. 7
• B. 10
• C. 14
• D. 16
• A. The object is at a distance less than \(f_o\) from the objective.
• B. The object is at a distance more than \(f_o\) and less than \(2f_o\) from the objective.
• C. The intermediate image is formed at a distance slightly more than \(f_e\) from the eyepiece.
• D. The intermediate image is formed at \(2f_e\) from the eyepiece.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A compound microscope works in two stages. The objective forms the first (intermediate) image of the object. The eyepiece then acts as a magnifier on this image.

Step 2: Nature of the intermediate image:
The object is kept just beyond the focus of the objective, between \(f_o\) and \(2f_o\). A convex lens then forms a real image, beyond \(2f_o\), which is inverted and larger than the object.

Step 3: Check the options:
(A) real, inverted and magnified: matches. TRUE.
(B) real, erect and magnified: a real image from a single convex lens is always inverted. FALSE.
(C) and (D) call the image virtual. A virtual image is formed only when the object is inside the focal length. In a microscope the object is outside it. FALSE.

Final Answer:
The intermediate image is real, inverted and magnified, option (A). \[ \boxed{\text{(A) real, inverted and magnified}} \]

Answer 2

Step 1: Understanding the Concept:
For the final image at infinity, the eyepiece must work as a simple magnifier. So the image made by the objective (the intermediate image) lies at the focal point of the eyepiece.

Step 2: Distance of the intermediate image:
The lens separation is \(15\ \text{cm}\) and \(f_e = 6.25\ \text{cm}\). So the intermediate image is at \(15 - 6.25 = 8.75\ \text{cm}\) from the objective. This is the image distance \(v_o = 8.75\ \text{cm}\).

Step 3: Use the lens formula for the objective:
\(\dfrac{1}{v_o} - \dfrac{1}{u_o} = \dfrac{1}{f_o}\) with \(f_o = 2.0\ \text{cm}\).
\[ \frac{1}{u_o} = \frac{1}{8.75} - \frac{1}{2.0} = 0.1143 - 0.5 = -0.3857\ \text{cm}^{-1} \]
\(u_o = -2.59\ \text{cm}\). So the object is \(2.59\ \text{cm}\) from the objective.

Step 4: Check the options:
(A) \(3.45\ \text{cm}\): this gives \(v = 2u/(u - 2) = 4.76\ \text{cm}\), which leaves \(15 - 4.76 = 10.2\ \text{cm}\) to the eyepiece, not \(6.25\ \text{cm}\). FALSE.
(B) \(5\ \text{cm}\): this gives \(v = 3.33\ \text{cm}\), leaving \(11.7\ \text{cm}\). FALSE.
(C) \(1.29\ \text{cm}\): this is less than \(f_o\), so the objective would form only a virtual image. FALSE.
(D) \(2.59\ \text{cm}\): this gives \(v = 8.78\ \text{cm}\) and the rest \(6.2\ \text{cm}\), equal to \(f_e\) within rounding. TRUE.

Final Answer:
The object must be about \(2.59\ \text{cm}\) from the objective, option (D). \[ \boxed{u_o \approx 2.59\ \text{cm}} \]

Answer 3

Step 1: Understanding the Concept:
The separation of the lenses is the image distance of the objective plus the distance of that image from the eyepiece. We are given the second distance, so we need the first from the objective lens formula.

Step 2: Image formed by the objective:
\(f_o = 0.8\ \text{cm}\), object distance \(u_o = -0.9\ \text{cm}\).
\[ \frac{1}{v_o} = \frac{1}{f_o} + \frac{1}{u_o} = \frac{1}{0.8} - \frac{1}{0.9} = 1.25 - 1.111 = 0.1389\ \text{cm}^{-1} \]
\(v_o = 7.2\ \text{cm}\).

Step 3: Add the eyepiece distance:
The intermediate image is \(2.27\ \text{cm}\) from the eyepiece. (Check: for a final image at the near point \(25\ \text{cm}\), \(1/u_e = -1/25 - 1/2.5 = -0.44\), so \(u_e = -2.27\ \text{cm}\).)
\[ L = v_o + |u_e| = 7.2 + 2.27 = 9.47\ \text{cm} \]

Step 4: Check the options:
(A) \(7.47\ \text{cm}\) and (B) \(8.47\ \text{cm}\) are not equal to \(7.2 + 2.27\), so they are FALSE.
(C) \(9.47\ \text{cm}\) matches. TRUE.
(D) \(10.47\ \text{cm}\) is one centimetre too large. FALSE.

Final Answer:
The separation between the two lenses is \(9.47\ \text{cm}\), option (C). \[ \boxed{L = 9.47\ \text{cm}} \]

Answer 4

Step 1: Understanding the Concept:
Magnifying power is \(m = m_o \times m_e\). For a final image at infinity, \(m_e = D/f_e\) with \(D = 25\ \text{cm}\). The objective magnification is \(v_o/|u_o|\).

Step 2: Image distance for the objective:
The intermediate image must lie at \(f_e = 10\ \text{cm}\) from the eyepiece. So \(v_o = 20 - 10 = 10\ \text{cm}\).

Step 3: Object distance and magnification of the objective:
\[ \frac{1}{u_o} = \frac{1}{v_o} - \frac{1}{f_o} = \frac{1}{10} - \frac{1}{2} = -0.4 \Rightarrow u_o = -2.5\ \text{cm} \]
\[ |m_o| = \frac{v_o}{|u_o|} = \frac{10}{2.5} = 4 \]

Step 4: Magnification of the eyepiece:
\[ m_e = \frac{D}{f_e} = \frac{25}{10} = 2.5 \]

Step 5: Total magnifying power:
\[ m = 4 \times 2.5 = 10 \]

Step 6: Check the options:
(B) 10 is the result. TRUE.
(C) 14 is what we would get for a final image at the near point: \(4 \times (1 + 25/10) = 4 \times 3.5 = 14\). The question says infinity, so it is FALSE.
(A) 7 and (D) 16 do not follow from any correct use of the data. FALSE.

Final Answer:
The magnifying power is 10, option (B). \[ \boxed{m = 10} \]

Answer 5

Step 1: Understanding the Concept:
Normal adjustment means the final image is at infinity. Then the eyepiece must receive parallel rays, so the intermediate image formed by the objective lies exactly at the focus of the eyepiece.

Step 2: What the objective does:
To give a real and magnified intermediate image, a convex lens needs an object between \(f_o\) and \(2f_o\). In a microscope the object is placed just beyond \(f_o\).

Step 3: Check statement (A):
If the object were closer than \(f_o\), the objective would give only a virtual image, which cannot be magnified again by the eyepiece in this way. So (A) is FALSE.

Step 4: Check statement (B):
The object lies a little more than \(f_o\) from the objective, which is more than \(f_o\) and less than \(2f_o\). So (B) is TRUE.

Step 5: Check statements (C) and (D):
In normal adjustment the intermediate image is at exactly \(f_e\) from the eyepiece, not slightly more. For a final image at the near point it would be slightly less than \(f_e\). So (C) is FALSE.
If the intermediate image were at \(2f_e\), the eyepiece would form a real image on its other side, not an image at infinity. So (D) is FALSE.

Final Answer:
The correct statement is (B). \[ \boxed{\text{(B)}} \]