AP Polycet (Polytechnic) 2025 Previous Question Paper with Answers And Model Papers With Complete Analysis

77) The focal length of a spherical mirror is 10cm. Its radius of curvature is:

A) 10cm
B) 5cm
C) 20cm
D) 0.2cm

View Answer

C) 20cm

Explanation: $f = \frac{R}{2} \Rightarrow R = 2f = 20 \, \text{cm}$
→ 20cm

78) An object placed between the principal focus and center of curvature of a convex lens forms an image:

A) beyond the center of curvature
B) at infinity
C) at the principal focus
D) between principal focus and center of curvature

View Answer

C) at the principal focus

Explanation:Object between F and 2F (focus and center of curvature) in convex lens → image is beyond 2F
→ beyond the center of curvature

79) The power of a lens is 4D.Its focal length is

A) 0.25cm
B) 2.5cm
C) 25cm
D) 0.025cm

View Answer

C) 25cm

Explanation:Power $P = \frac{100}{f(\text{cm})} \Rightarrow f = \frac{100}{4} = 25 \, \text{cm}$
→ 25cm

80) If the height of the image is equal to the height of an object placed near a spherical lens, then the magnification m is

A) less than 1
B) greater than 1
C) equal to 1
D) equal to zero

View Answer

C) equal to 1

Explanation:Image height = object height → magnification = 1
→ equal to 1

81) An object is placed at a distance of 30cm from a concave lens of focal length 20cm.
The image distance is:

A) 75cm
B) 60cm
C) 12cm
D) 50cm

View Answer

C) 12cm

Explanation:Use lens formula:
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}, \quad f = -20\,\text{cm},\, u = -30\,\text{cm} \Rightarrow \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = -\frac{1}{20} - \frac{1}{30} = -\frac{1}{12} \Rightarrow v = -12\,\text{cm}$
Image distance = 12cm
→ 12cm

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