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1.	It was a bright, sunny day in the morning when the classes started in the school. Suddenly dark clouds appeared in the sky. Due to this the classroom became quite dark. The teacher asked a student to switch on all the lights in the classroom. (a) Why did the teacher ask for all the lights to be switched on ? (b) What was the size of the pupil of the eyes of the students : (i) before the clouds appeared ? (ii) when classroom became quite dark ? (iii) when the lights were switched on ? (c) In which of the above mentioned three situations, the students will feel glare in their eyes and why ? (d) What values are exhibited by the teacher in this case ?
Answer» (a) The teacher asked all the lights in the classroom to be switched on because seeing the blackboard writing in dim light put a lot of strain on the eyes of the students. (b) (i) Before the clouds appeared, there was sufficient light in the classroom due to which the size of pupil of eyes was small. (ii) When classroom became quite dark, the light was dim, due to which the size of pupil of eyes was very large. (iii) When the lights were switched on, the light was bright and hence the size of pupil of eyes became small again. (c) The students will feel glare in their eyes when the lights are switched on in the dark classroom. This is because the size of pupil of eye is large in the dark room. So, when lights are switched on, then suddenly too much light enters the eyes (due to large size of pupil) causing the glare in the eyes. (d) The values exhibited by the teacher are (i) General awareness or Knowledge (that it is necessary to study in sufficient light to keep the eyes healthy, and (ii) Concern for the students (that studying in dim light can harm the eyes of students).

2.	The near point of a hypermetropic eye is 1 m. What is the nature and power of the lens required to correct this defect ? (Assume that the near point of the normal eye is 25 cm).
Answer» The eye defect called hypermetropia is corrected by using a convex lens. So, the person requires convex lens spectacles. We will first calculate the focal length of the convex lens required in this case. This hypermetropic eye can see the nearby object kept at 25 cm (at near point of normal eye ) clearly if the image of this object is formed at its own near point which is 1 metre here. So, in this case : Object distance, u = -25 cm `" "` (Normal near point) Image distance, v = -1m `" "` (Near point of this defective eye) = - 100 cm And, Focal length, `f = ?" "` (To be calculated) Putting these value in this lens formula, `(1)/(v)-(1)/(u) = (1)/(f)` We get : `(1)/(-100) - (1)/(-25) = (1)/(f)` or `- (1)/(100) + (1)/(25) = (1)/(f)` `(-1 + 4)/(100) = (1)/(f)` `(3)/(100) = (1)/(f)` `f = (100)/(3)` `f = 33.3 cm` Thus, the focal length of the convex lens required is + 33.3 cm. We will now calculate the power. Please note that 33.3 cm is equal to `(33.3)/(100)` m or 0.33 m. Now, Power, `P = (1)/(f ("in metres"))` `= (1)/(+ 0.33)` `= + (100)/(33)` `= + 3.0 D` So, the power of convex lens required is + 3.0 dioptres. Myopia and hypermetropia are the two most common defects of vision (or defects of eye). We will now study another defect of vision which occurs in old age. It is called presbyopia.

3.	Why do stars twinkle ?A.B.C.D.
Answer» See page 290 of this book.

4.	Explain why the planets do not twinkle .A.B.C.D.
Answer» See page 291 of this book.

5.	Why does the sum appear reddish in the morning ?A.B.C.D.
Answer» See page 296 of this book.

6.	Sanjay was going to his office in his car. While driving his car, Sanjay saw a man behind him on a motorcycle through his rear-view mirror. A woman was also sitting behind the man on the motorcycle. Through his rear-view mirror, Sanjay noticed that the saree of the woman was almost touching the spokes of motorcycle wheel. He signalled the motorcyclist to stop and alerted the woman. She tied her saree properly and thanked Sanjay for the alert. (a) What kind of mirror is used by Sanjay as rear-view mirror ? (b) State two characteristics of the image formed by such a rear-view mirror ? (c) What could have happened if Sanjay had not alerted the woman sitting at the back of motorcycle ? (d) What values are displayed by Sanjay in this incident ?
Answer» (a) Convex mirror. (b) The characteristics of the image formed by rear-view mirror (which is a convex mirror), are (i) virtual and erect, and (ii) diminished (smaller than the object). (c) If Sanjay had not alerted the woman sitting at the back of motorcycle, then her saree could get entangled in the spokes of moving motorcycle wheel and cause a serious accident. (d) The values displayed by Sanjay in this incident are (i) Vigilant (because he kept careful watch for possible danger around him) (ii) Concern about others (here he was concerned about the safety of the woman), and (iii) Responsible citizen (because he stopped the motorcycle and alerted the woman about impending danger to her life).

7.	Anahat is an eight year old girl who has just begun to learn swimming. At the moment she is in the swimming pool. Suddenly, Anahat finds that her earring has fallen into the water. It appears to her that the earring has not gone too deep. So, she begins to go further into swimming pool to retrieve her earring. Her trainer, who has been watching her movements, does not allow Anahat to go further and pulls her out of the swimming pool quickly. (a) Why does it appear to Anahat that her earring has not gone too deep in water ? Name the phenomenon involved. (b) Why does the trainer not allow Anahat to go further in the swimming pool ? (c) Give one example of the effect of phenomenon involved in the above episode in which a glass slab is used. (d) What values does the trainer show in the above episode ?
Answer» (a) The Phenomenon involved is refraction of light. Due to refraction of light coming out through the water of swimming pool into air, the bottom of swimming pool appears raised. In other words, the swimming pool appears to be less deep to Anahat than it really is. Due to this, Anahat thinks that her earring has not gone too deep in water and tries to retrieve it. (b) The trainer does not allow Anahat to go further in the swimming pool because the water there is much deeper than it appears to be and there is a risk of drowning (c) Due to refraction of light, when a glass slab is placed over some printed matter, the letters appear to be raised when viewed from top. (d) The trainer shows the values of (i) Sense of duty (ii) Concern for the safety of children, and (iv) Timely action.

8.	Why is a normal eye not able to see clearby the objects placed closer than 25cm .A.B.C.D.
Answer» A normal eye can see nearby objects because the ciliary muscles make the eye - lens more convex (or thick ) thereby increasing its converging power. This increased converging power of eye-lens can make a normal eye see objects clearly only when placed up to 25 cm from the eye. However, when objects are placed closer than 25cm, then the ciliary muscles cannot make the eye-lens more convex to increase its converging power further So as to focus them sharply on the retina. Due to this, such objects are not seen clearly. In other words, a normal eye is not able to see clearly the objects placed closer than 25cm because all its power of accommodation (or ability to increase converging power of eye - lens by making it more convex ) has already been exhausted .

9.	What is meant by the power of accommodation of the eye ?A.B.C.D.
Answer» the ability of an eye to focus the distant objects as well as the nearby in the retina by changing the focal length (or converging power ) of its lens is called accommodation .

10.	A person with a myopic eye connot see object beyond 1.2 m distinctly what should be the type of the corretive lens used to restore proper vision ?A.B.C.D.
Answer» The proper vision of a mypic eye can be restored by using concave lens of suitable power .

11.	The far point of a myopic person is 80 cm in front of the eye. What is the nataure and power of the lens required to correct the problem ?A.B.C.D.
Answer» see sample problem on page 275 of this book.

12.	What are the far point and near point of the human eye with normal vision ?A.B.C.D.
Answer» The far point of human eye with normal vision is at infinity and the near point is at a distance of 25 centimetres from the eye .

13.	(a) Make a diagram to show how hypermetropia is corrected (b) The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect ? Assume that the near point of the normal eye is 25 cm .A.B.C.D.
Answer» (a) see figure 13(c) on page 276 of this book . (b) see sample problem on page 277 of this book .