Home
Class 10
PHYSICS
" Object " " at " 2F(1) : " Image "" at ...

`" Object " " at " 2F_(1) : " Image "" at "2F_(2) :: " Object " " at "F_(1)` : ……….

Text Solution

Verified by Experts

Image at infinity : In convex lens when the object is at the focus, a very large, real and inverted image will be formed at infinity.
Promotional Banner

Topper's Solved these Questions

  • LENSES

    CHETAN PUBLICATION|Exercise Match the columns|6 Videos
  • LENSES

    CHETAN PUBLICATION|Exercise .True of .False..|8 Videos
  • LENSES

    CHETAN PUBLICATION|Exercise Assignment|14 Videos
  • HEAT

    CHETAN PUBLICATION|Exercise ASSIGNMENT - 5 (Answer the following: (Any 1))|3 Videos
  • MODEL ACTIVITY SHEET

    CHETAN PUBLICATION|Exercise SOLVE THE FOLLOWING QUESTIONS : (ANY 1)|9 Videos

Similar Questions

Explore conceptually related problems

If an object is placed between F_(1)" and " 2F_(1) of a convex lens, then nature of the image formed is ………….

Consider f" ":" "{1," "2," "3}->{a ," "b ," "c} given by f(1)" "=" "a , f(2)" "=" "b and f(3)" "=" "c . Find f^(-1) and show that (f^(-1))^(-1)=" "f .

If y=mx+c " and " f(0)=f'(0)=1, " then " f(2) is

Let f be the fundamental frequency of the string . If the string is divided into three segments l_(1) , l_(2) and l_(3) such that the fundamental frequencies of each segments be f_(1) , f_(2) and f_(3) , respectively . Show that (1)/(f) = (1)/(f_(1)) + (1)/(f_(2)) + (1)/(f_(3))

A ray diagram for object position at 2F_(1) for a convex lens.

A ray diagram for object positioned at F_(1) for a convex lens.

An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length f_(0) of the objective and f_(e) of the eyepiece are respectively

A function f [-3, 7] rarr R is defined as follows f(x) = {{: (4x^(2) - 1, " " -3 le x lt 2 ),(3x -2, " " 2 le x le 4 ),(2x -3 , " " 4 lt x lt 7):} Find (i) f (5) + f(6) (ii) f(1) - f(-3) (iii) f(-2)-f(4) (iv) (f(3)+ f(-1))/(2 f(6) - f(1))

If y = mx + c and f(0) = f'(0) =1 , then f(2) is …......... .