Vidaara.orgClass 10 · Physics
CodeVID-P10-02-DEF-01
Defects of Vision & Correction — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for numericals. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
Hypermetropia is corrected using a
- A.concave lens
- B.convex lens
- C.cylindrical lens
- D.plane mirror
2.
The SI unit of power of a lens is the
- A.metre
- B.dioptre
- C.watt
- D.newton
3.
A convex lens has power
- A.negative
- B.positive
- C.zero
- D.infinite
4.
Loss of accommodation with old age is called
- A.myopia
- B.hypermetropia
- C.presbyopia
- D.cataract
5.
Power of a lens of focal length $+2\,\text{m}$ is
- A.$+0.5\,\text{D}$
- B.$+2\,\text{D}$
- C.$-0.5\,\text{D}$
- D.$+1\,\text{D}$
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Write the cause of myopia.
7.
Calculate the power of a convex lens of focal length $20\,\text{cm}$.
8.
Why can a cataract not be corrected by spectacles?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
The far point of a myopic eye is $1.5\,\text{m}$. Find the focal length and power of the correcting lens.
10.
Differentiate between myopia and hypermetropia in terms of image position and corrective lens.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
A hypermetropic person has a near point of $75\,\text{cm}$. Find the power of the lens to read at $25\,\text{cm}$, and draw a ray diagram of the correction.
Answer Key
Section A — Multiple Choice Questions
- (B) convex lens
- (B) dioptre
- (B) positive
- (C) presbyopia
- (A) $+0.5\,\text{D}$
Section B — Short Answer (2 marks)
- Eyeball too long or eye lens too curved, so image of distant object forms in front of the retina.
- $P=1/0.2=+5\,\text{D}$.
- The lens itself turns cloudy; light cannot pass clearly, so only surgery helps.
Section C — Short Answer (3 marks)
- $f=-1.5\,\text{m}$, $P=-0.67\,\text{D}$ (concave).
- Myopia: image in front of retina, concave lens; hypermetropia: image behind retina, convex lens.
Section D — Long Answer (5 marks)
- $u=-0.25\,\text{m}$, $v=-0.75\,\text{m}$; $1/f=1/-0.75-1/-0.25=2.67$; $P=+2.67\,\text{D}$ (convex).
Generated by Vidaara.org · Assignment VID-P10-02-DEF-01 · vidaara.org