TOP 5 QUESTION ASKED IN CBSE FROM CHAPTER LIGHT (CLASS 10)
Chapter: light-reflection and refraction
Ques 1. An object is kept in front of concave mirror of focal length 15cm. The image formed is three times the size of object. Calculate two possible distances of object from the mirror. (फोकल लम्बाई 15 सेंटीमीटर के अवतल दर्पण के सामने एक वस्तु रखी जाती है। बनाई गई छवि ऑब्जेक्ट के आकार के तीन गुना है। दर्पण से ऑब्जेक्ट की दो संभावित दूरी की गणना करें). 3marks
Ques 2. What is power of lens?? If power is +2.5D. What kind of lens is it and what is it's focal length.(लेंस की क्षमता क्या है ?? अगर क्षमता + 2.5 डी है। यह किस प्रकार का लेंस है और इसकी फोकल लम्बाई क्या है।). 2marks
Ques 3. Draw ray diagram of concave mirror when object is placed between c and f . Write nature of image(सी और एफ के बीच ऑब्जेक्ट रखा जाता है तब अवतल दर्पण के रे आरेख ड्रा। छवि की प्रकृति लिखें). 3marks
Ques 4. Write three difference between real and virtual images(वास्तविक और आभासी छवियों के बीच तीन अंतर लिखें). 2marks
Ques 5. What is law of reflection and law of refraction. Draw diagram to differentiate reflection and refraction(परावर्तन और अपवर्तन के कानून क्या है।परावर्तन और अपवर्तन को अलग करने के लिए चित्र बनाएं). 3marks
This is Raj Q-1 answer first possible distance of object from concave mirror is -10cm and second possible distance is -20cm
ReplyDeleteThis is Raj Q2 answer- power of lens is reciprocal of focal length (in metre) and if power of lens is +2.5D so it is convex lens and focal length of this lens is 40cm
ReplyDeleteQ-3 answer - when the object is placed between c and f then image is formed beyond c and nature of image is real, inverted and enlarged
ReplyDeleteQ-4 answer- Real image : -
ReplyDelete(a) The light rays actually meet.
(b) can be obtained on screen
(c) it is inverted
Virtual image:-
(a) The light rays appears to meet
(b) can't obtained on screen
(c) it is erect
Q5-answer- Law of reflection
ReplyDelete(a) Angle of incidence is equal to Angle of reflection
(b) the incident ray , reflected ray and the normal all lie in same plane
Law of refraction
(a) the incident ray , the refracted ray and normal all lie in same plane.
(b) Snell's law :- The ratio of sin function of angle of incidence to the sin function of angle of refraction is constant in same medium
sin i / sin r = constant
Aman give me solution of Q1
ReplyDeleteI am Aman, solution of Q1
ReplyDeleteHey aman, As in ques 1 there is concave mirror so f=-15cm
DeleteAnd m=h'/h= -v/u=3
So we can say v= -3u so by mirror formula 1/f = 1/v + 1/u
1/f = 1/-3u + 1/u
1/f = -1+3/3u
1/f = 2/3u
1/-15 = 2/3u and u = -15 ×2/3
u= -10cm it means object is at 10 cm from mirror
As in question there is no mention of nature of image so we can take magnification = -3
It means m = h'/h = -v/u =-3
Now v = +3u so by mirror formula 1/f =1/v + 1/u
1/f = 1/3u + 1/u
1/f = 4/3u so u = 4×f/3 = 4×-15/3
u = -20cm
So there will be two possible distance I.e -10cm and -20 cm
Size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/ 3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?
ReplyDelete