Question

Find the position of the object when an image is formed in front of a mirror on a screen at a distance of 30cm from the mirror. The focal length of the mirror is 10cm. Calculate magnification​

Answer 1

AnswEr :

Given that ,

The two resistor 35 ohm and 14 ohm are connected in parrallel combination with 7 v battery

We know that , the equivalent resistance in parrallel combination is given by

$\large \rm \fbox{ \frac{1}{R } = \frac{1}{R_{1}} + \frac{1}{R_{2}} + ... + \frac{1}{R_{n}} }$

Thus ,

$\sf \mapsto \frac{1}{R} = \frac{1}{35} + \frac{1}{14} \\ \\ \sf \mapsto \frac{1}{R} = \frac{14 + 35}{490} \\ \\ \sf \mapsto R = \frac{490}{49} \\ \\ \sf \mapsto R =10 \: \: ohm$

$\therefore \sf \underline{The \: equivalent \: resistance \: is \: 10 \: ohm}$

We know that ,

$\large \rm \star \: \: \fbox{V = IR} \: \: \{ \because \: Ohm's \: law\}$

Thus , the current in each resistor will be

7 = I × 35

I = 0.2 amp

And

7 = I × 14

I = 0.5 amp

$\therefore \sf \underline{The \: current \: in \: each \: resistor \: are \: 0.2 \: amp \: and \: 0.5 \: amp}$

Now , the net current will be

7 = I × 10

I = 0.7 amp

$\therefore \sf \underline{The \: net \: current \: is \: 0.7 \: amp }$

Answer 2

search-icon-header

Search for questions & chapters

search-icon-image

Class 12

>>Physics

>>Ray Optics and Optical Instruments

>>Spherical Mirrors

>>An object is placed at a distance of 30

Question

Bookmark

An object is placed at a distance of 30 cm. from a concave mirror. The magnification produced is 1/2. where should the object be placed to get the magnification of 1/3.

Hard

Solution

verified

Verified by Toppr

U=−30

m=

2

1

we know that

m= U−V

so V = 15 f=10 for m= 31

U−V =31

V= 3−V

f1 = V1 + U1

101 = −U3 + U1

U = -40

Therefore for magnification to be 31

the object must be placed 40cm upon in front of the mirror.