Question

Magnification of convex lens when objects is placed at 2F is​

Answer 1

Answer:

$hey \: mate$

Explanation:

Case 2: The object is located at 2F

The image dimensions are equal to the object dimensions. A six-foot tall person would have an image that is six feet tall; the absolute value of the magnification is exactly 1. Finally, the image is a real image. Light rays actually converge at the image location.

plz mark as a brainlist

Answer 2

$\rule{200}2$

$\large\underline{\underline{\red{\purple{\maltese} \: \textsf{ Case : When the object is at 2F :- }}}}$

When the object is placed at 2F in front of a convex lens , then the image is formed at $\sf 2F_2$. The size of the image is equal to the size of the object but the image formed is inverted . So according to the sign convention the height of the image will be negative . Let the height of the image be h , then the height of image = -h . So ,

$\sf:\implies \pink{Magnification = \dfrac{ Height \ of \ the \ image }{Height \ of \ the \ object }}\\\\\sf:\implies Magnification = \dfrac{-h}{h} </p><p>\\\\\sf:\implies Magnification = \dfrac{-1}{1}\\\\\sf:\implies\boxed{\pink{\mathfrak{ Magnification = -1}}}$

$\underline{\blue{\sf Hence \ the \ magnification \ is \ \textsf{\textbf{-1 }}. }}$

$\rule{200}2$

$\underline{\underline{\red{\purple{\maltese} \: \textsf{ Some More Information:- }}}}$

• Sign Convention :-

$\blue{\longmapsto}$ The distances measured along the direction of incident ray are taken as positive and that opposite to the direction of incident rays are taken as negative.

$\blue{\longmapsto}$ All the distances parallel to the principal axis are measured from the optical centre.

$\blue{\longmapsto}$ All the distances measured perpendicular to and above the principal axis are taken positive while those measured below it are taken as negative.

$\blue{\longmapsto}$ Focal length is taken positive for convex lens while negative for concave lens object distance is always taken as negative for all objects.

$\setlength{\unitlength}{1 cm}\begin{picture}(12,12)\thicklines\put(3,0){\vector(-1,0){3}}\put(3.001,0){\vector(1,0){3}}\put(3,0.01){\vector(0,-1){3}}\put(3,-0.01){\vector(0,1){3}}\put(1,1.5){\vector(1,0){1.5}}\put(-2,0.5){\sf Principal \ Axis } \put(1,1.9){\sc Light }\put(0,-0.3){\sf -ve } \put(6,-0.3){\sf +ve} \put(3,3.3){\sf +ve}\put(3,-3.3){\sf -ve}\put(3.2,-0.35){\sf O}\end{picture}$

$\rule{200}2$