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: In an astronomical telescope, the objective lens receives nearly parallel rays coming from a very distant object. The objective lens forms a real, inverted and diminished image at its focal plane. This real image acts as the object for the eyepiece. The eyepiece behaves like a simple microscope and produces a virtual, enlarged image for comfortable viewing.

Step 1: Nature of image formed by objective lens. Since the object is at a very large distance, the rays reaching the objective lens are almost parallel. The objective lens therefore forms the image at its principal focus. This image is:
• Real
• Inverted
• Diminished

Step 2: Nature of image formed by eyepiece. The real image formed by the objective serves as an object for the eyepiece. The eyepiece magnifies this image and produces a final image which is:
• Virtual
• Enlarged
• Inverted with respect to the original object Therefore, \[ \boxed{\text{Objective image = Real}} \] and \[ \boxed{\text{Eyepiece image = Virtual}} \] Hence, \[ \boxed{\text{Correct Option (B)}} \]

Answer 2

Concept: The magnifying power of an astronomical telescope in normal adjustment is \[ M=\frac{f_o}{f_e} \] where \[ f_o \] is the focal length of the objective lens and \[ f_e \] is the focal length of the eyepiece.

Step 1: Examine the expression for magnifying power. From \[ M=\frac{f_o}{f_e} \] it is clear that magnifying power depends upon:
• Focal length of objective lens
• Focal length of eyepiece

Step 2: Role of aperture. The aperture determines:
• Brightness of image
• Light gathering power
• Resolving power However, it does not appear in the expression for magnification. Therefore, magnification is independent of aperture. Hence, \[ \boxed{\text{Correct Option (D)}} \]

Answer 3

Step 1: Recall the condition for normal adjustment. In normal adjustment, \[ \text{Distance between objective and eyepiece} = f_o+f_e. \] It is exactly equal to the sum of the focal lengths.

Step 2: Check each statement. Statement (A): \[ f_o>f_e \] This is true because telescope objectives have large focal lengths. Statement (B): \[ M=\frac{f_o}{f_e} \] Increasing \(f_o\) increases \(M\). Hence true. Statement (C): It says distance between lenses is more than \[ (f_o+f_e). \] This is incorrect because in normal adjustment the separation equals \[ f_o+f_e. \] Statement (D): Increasing \(f_e\) decreases \[ M=\frac{f_o}{f_e}. \] Hence true. Therefore, \[ \boxed{\text{Correct Option (C)}} \]

Answer 4

Concept: For normal adjustment of an astronomical telescope, \[ L=f_o+f_e. \]

Step 1: Substitute the given values. Given, \[ f_o=80\,\text{cm} \] and \[ f_e=4\,\text{cm}. \] Hence, \[ L=f_o+f_e \] \[ L=80+4 \] \[ L=84\,\text{cm}. \]

Step 2: Write the final answer. Therefore, \[ \boxed{L=84\,\text{cm}} \] Hence, \[ \boxed{\text{Correct Option (A)}} \]

Answer 5

Concept: The magnifying power of an astronomical telescope in normal adjustment is \[ M=\frac{f_o}{f_e}. \]

Step 1: Substitute the given focal lengths. Given, \[ f_o=80\,\text{cm} \] and \[ f_e=4\,\text{cm}. \] Therefore, \[ M=\frac{80}{4}. \] \[ M=20. \]

Step 2: Interpret the result. The telescope makes the distant object appear twenty times larger in angular size than that seen by the unaided eye. Thus, \[ \boxed{M=20} \] Hence, \[ \boxed{\text{Correct Option (D)}} \]