The shortest distance between an object and its real image formed by c

1.	The shortest distance between an object and its real image formed by converging lens of focal length , 10 cm is _______
Answer» Minimum distance between object and real image in the converging lens is 40 cm. Explanation: Let, u=object distance. v= image distance. f= focal length of the lens. x= distance between object and image. We already know that to obtain real image v must be positive. Now we should use the lens formula, 1/v-1/u=1/f Now, u = x - v. Again we should put the appropriate value according toCartesian system of sign convention u is negative, we get, 1/v+1/(x-v)=1/f. ⇒ (x-v+v)/[v(x-v)] = 1/f. ⇒ xf/(vx - v²) = 1 ⇒ xf = vx - v². Then, v²- vx + xf = 0 Let's take the condition for real root of this quadratic equation in v. Now, the roots are: v=(1/2)[x+/- √(x²–4xf)]. For v to be real, x²> 4xf or, x > 4f So the minimum distance between object and real image is 4 x 10 = 40 cm.

The shortest distance between an object and its real image formed by converging lens of focal length , 10 cm is _______

Answer»

Minimum distance between object and real image in the converging lens is 40 cm.

Explanation:

Let,
u=object distance.
v= image distance.
f= focal length of the lens.
x= distance between object and image.

We already know that to obtain real image v must be positive.

Now we should use the lens formula,

1/v-1/u=1/f

Now, u = x - v.
Again we should put the appropriate value according toCartesian system of sign convention u is negative, we get,

1/v+1/(x-v)=1/f.
⇒ (x-v+v)/[v(x-v)] = 1/f.
⇒ xf/(vx - v²) = 1
⇒ xf = vx - v².

Then,

v²- vx + xf = 0

Let's take the condition for real root of this quadratic equation in v.

Now, the roots are:

v=(1/2)[x+/- √(x²–4xf)].

For v to be real, x²> 4xf

or, x > 4f

So the minimum distance between object and real image is 4 x 10 = 40 cm.

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