Question

The images formed by the objective lens and the eyepiece are respectively :

• A. virtual, real
• B. real, virtual
• C. virtual, virtual
• D. real, real
• A. colour of light
• B. focal length of objective lens
• C. focal length of eyepiece
• D. apertures of objective lens and eyepiece
• A. The focal length of objective lens ($f_o$) is larger than the focal length of eyepiece ($f_e$).
• B. Its magnifying power can be increased by increasing the focal length of objective lens ($f_o$).
• C. The distance between two lenses is more than ($f_o + f_e$).
• D. The magnifying power can be decreased by increasing the focal length of eyepiece.
• A. 84 cm
• B. 76 cm
• C. 20 cm
• D. 320 cm
• A. 320
• B. 84
• C. 76
• D. 20

Answer 1

The Correct Option is B

Concept:
• A refracting astronomical telescope utilizes two convex lenses to observe very distant objects.
• The objective lens has a large focal length and a large aperture to gather maximum light from the distant astronomical body.
• The eyepiece has a comparatively smaller focal length and smaller aperture, acting essentially as a simple magnifier for the image produced by the objective.

Step 1: Analyze the image formed by the objective lens
The object being viewed (like a star or a planet) is situated practically at infinity.
When parallel rays of light from this distant object enter the objective lens, they converge at the focal plane of the objective.
Because the light rays actually intersect to form this intermediate image, it is a real image.
Furthermore, this real image is inverted and highly diminished compared to the actual object size.

Step 2: Analyze the image formed by the eyepiece
This intermediate real image acts as the optical object for the secondary lens, the eyepiece.
In a properly adjusted telescope (especially for normal adjustment), this intermediate object is positioned slightly within or exactly at the focal length of the eyepiece.
When an object is placed between the optical center and the principal focus of a convex lens, the lens produces an image that is virtual, erect (with respect to the intermediate object), and highly magnified.
Since the rays diverge after passing through the eyepiece, they only appear to meet when produced backward, confirming the final image is purely virtual.

Step 3: Conclusion
Summarizing the optical actions of both lenses:
The objective lens forms a real image.
The eyepiece takes this real image and forms a virtual final image.
Therefore, the sequence of images is real, followed by virtual.

Step 4: Final Answer
The correct option perfectly matching this derived sequence is (B).

Answer 2

Concept:
• The magnifying power (or angular magnification) of a telescope is the ratio of the angle subtended at the eye by the image to the angle subtended by the object.
• In normal adjustment (when the final image is at infinity), the magnifying power $m$ is given by $m = \frac{f_o}{f_e}$.
• In near point adjustment (when the final image is at the least distance of distinct vision $D$), $m = \frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right)$.

Step 1: Analyze dependence on focal lengths
From the standard formulas mentioned in the concepts, it is mathematically evident that magnification heavily depends on both $f_o$ (focal length of objective) and $f_e$ (focal length of eyepiece).
An increase in the objective's focal length directly increases magnification, while an increase in the eyepiece's focal length decreases it.
Thus, options (B) and (C) are directly involved in the magnification formula.

Step 2: Analyze dependence on color of light
According to Lens Maker's Formula, the focal length of a lens is inversely related to $(n - 1)$, where $n$ is the refractive index of the material.
Cauchy's equation tells us that the refractive index $n$ varies strictly with the wavelength (or color) of light.
Therefore, a change in the color of light leads to a change in the refractive index, which in turn alters the focal lengths $f_o$ and $f_e$.
Since the focal lengths change based on color, the overall magnifying power is implicitly dependent on the color of light.

Step 3: Analyze dependence on apertures
The aperture of a lens refers to its effective diameter or the light-gathering area.
While a larger aperture for the objective lens dramatically improves the resolving power and the brightness of the resulting image, it completely fails to alter the geometric focal lengths.
Since the aperture size does not appear in the angular magnification equations at all, the magnifying power remains completely independent of the apertures.

Step 4: Conclusion
The magnifying power depends on focal lengths and color, but definitely not on the physical aperture sizes of the lenses.
Hence, the correct option is (D).

Answer 3

Concept:
• An astronomical telescope is designed to view extremely distant objects, necessitating specific lens configurations.
• To achieve high magnification, the formula $m = \frac{f_o}{f_e}$ dictates that $f_o \gg f_e$.
• The tube length of the telescope ($L$) represents the separation between the objective lens and the eyepiece.

Step 1: Evaluate Statement (A)
For a functional astronomical telescope, the objective must gather abundant light and the system must provide angular magnification.
This fundamentally requires the focal length of the objective ($f_o$) to be significantly greater than the focal length of the eyepiece ($f_e$).
Therefore, statement (A) is completely correct in its assertion.

Step 2: Evaluate Statements (B) and (D)
Using the fundamental relation for magnifying power in normal adjustment: $m = \frac{f_o}{f_e}$.
It is mathematically obvious that magnification is directly proportional to $f_o$. Increasing $f_o$ will indeed increase the magnifying power, making statement (B) correct.
Conversely, magnification is inversely proportional to $f_e$. Increasing $f_e$ will effectively decrease the magnifying power, making statement (D) entirely correct.

Step 3: Evaluate Statement (C)
The physical distance between the two lenses is known as the tube length $L$.
When the telescope is adjusted for normal vision (final image at infinity), the intermediate image forms exactly at the focal points of both lenses, giving $L = f_o + f_e$.
When adjusted for near point vision (final image at distance $D$), the intermediate image forms closer to the eyepiece, yielding $L = f_o + u_e$, where $u_e < f_e$.
In both typical adjustments, the distance $L$ is either exactly equal to or less than ($f_o + f_e$).
It is never more than ($f_o + f_e$).
Thus, statement (C) is the logically incorrect statement.

Step 4: Conclusion
Since the question asks for the incorrect statement, option (C) fits perfectly.

Answer 4

Concept:
• When a telescope is in "normal adjustment," it implies that the final image is being formed at optical infinity.
• This is the most relaxed viewing state for the human eye.
• To achieve this, the real image formed by the objective lens must fall precisely on the primary focal plane of the eyepiece.

Step 1: Identify the given parameters
From the text of the problem, the focal length of the objective lens is given as $f_o = 80 \text{ cm}$.
The focal length of the eyepiece is given as $f_e = 4 \text{ cm}$.

Step 2: Apply the normal adjustment condition
In normal adjustment, the principal focus of the objective lens perfectly coincides with the principal focus of the eyepiece.
The physical distance separating the optical centers of the two lenses is referred to as the length of the telescope tube, denoted by $L$.
The geometric relationship for this specific alignment is given simply by the sum of their individual focal lengths:
\[ L = f_o + f_e \]

Step 3: Perform the final calculation
Substitute the given numerical values into the tube length equation:
\[ L = 80 \text{ cm} + 4 \text{ cm} \]
\[ L = 84 \text{ cm} \]
This means the lenses must be separated by exactly 84 cm for parallel light to emerge from the eyepiece.

Step 4: Conclusion
The computed distance is 84 cm, which corresponds exactly to option (A).

Answer 5

Concept:
• Magnifying power (or angular magnification) is a measure of how much larger the object appears when viewed through the telescope compared to the naked eye.
• For an astronomical telescope set to normal adjustment, the final image is projected at infinity.
• The standard formula relating magnifying power $m$ to the lens parameters is $m = \frac{f_o}{f_e}$.

Step 1: Extract the required parameters
Referring back to the data provided in part (iv)(a):
The objective lens focal length is given by $f_o = 80 \text{ cm}$.
The eyepiece focal length is provided as $f_e = 4 \text{ cm}$.

Step 2: Apply the magnification formula
For viewing an object at infinity with the final image also placed at infinity (normal adjustment), the magnification is solely determined by the ratio of the focal lengths.
Utilize the formula:
\[ m = \frac{f_o}{f_e} \]

Step 3: Execute the arithmetic
Substitute the known values into the ratio:
\[ m = \frac{80 \text{ cm}}{4 \text{ cm}} \]
Since both quantities are in centimeters, the units cancel out elegantly, yielding a dimensionless number.
\[ m = 20 \]
This calculated value means the telescope makes distant objects appear 20 times larger in angular terms.

Step 4: Conclusion
The calculated magnifying power is 20, which aligns precisely with option (D).