Question

What should be a distance of an object from convex lense of focal length 10cm in order to produce an erect image twice as large as object? ​

Answer 1

Answer:

opposite

Explanation:

due to some critical thinking

Answer 2

★ Solution:-

• focal length = + 10 cm

focal length of a convex lens is always positive

• Let the height of the object be ho and the height of the image be hi

then , hi = 2 ho

As the image formed is erect the object will be virtual and height will be positive

By using the magnification formula ,

$\longrightarrow\mathtt{m = \dfrac{v}{u}= \dfrac{h_i}{h_o} }$

here ,

• v = image distance
• u = object distance
• hi = height of image
• ho = height of object

$\longrightarrow\mathtt{v= \dfrac{2h_o}{h_o}\times u }$

$\longrightarrow\mathtt{v= 2 u }$

Now let's find u by substituting the value of v in the lens formula,

$\longrightarrow\mathtt{\dfrac{1}{f}= \dfrac{1}{v} - \dfrac{1}{u}}$

$\longrightarrow\mathtt{\dfrac{1}{f}= \dfrac{1}{2u} - \dfrac{1}{u}}$

$\longrightarrow\mathtt{\dfrac{1}{f}= \dfrac{1-2}{2u}}}$

$\longrightarrow\mathtt{\dfrac{1}{f}= \dfrac{-1 }{2u}}}$

$\longrightarrow\mathtt{u = \dfrac{-f }{2} }}$

Now let's substitute the value of f in the above equation

$\longrightarrow\mathtt{u = \dfrac{-10 }{2} }}$

$\longrightarrow \boxed{\mathtt{u = - 5 cm}}}$

(-ve sign denotes that the object is placed in front of the mirror as distances are measured from the optical center of the lens )

The object is placed at a distance of 5 cm from the concave lens