A certain far - sighted person cannot see object that are closer to the eye than 50 cm clearly. Determine the power of the converging lens which will enable him to see at 25 cm
0.06 D
0.02 D
0.03 D
0.04 D
no option is correct
Explanation
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Since he is long sighted, he will use a convex lens which produce real images, and note that abnormal near points and far points have negative distances. we have
F = uv/u+v
f = -1250/25-50
f= -1250/-25
f = 50cm
We all know that power of a lens is focal length in metres, so we have
P = 100/f
P= 100/50
P = 2.0D
Without any doubt, it means that the correct answer is E.
The selected answer is wrong:
1/f= 1/u + 1/u
f=?
u=25cm
v=-50cm
1/f=1/25+1/-50
1/f=(2-1)/50
Cross multiply
f=50cm
Power is reciprocal of focal length
p=1/f
p=1/50 =0.02
P= 0.02 D
REF: check any tb
the right formula to use is (1/-v)-(-1/u)=1/f
-1/50+1/25=1/f
1/25-1/50=1/f
2-1/50=1/f
1/50=1/f
f=50
d=1/50
d=0.02. B although the forcal lenght should have been in meter not centimeter, but if u do dat the answer will be=2D which is not given. so the correct answer is B. 0.02D.
Power is the rate at which work is done or energy is transferred. Here are some key formulas related to power:
### **General Power Formula**
\[
P = \frac{W}{t}
\]
Where:
- \( P \) = Power (Watts, W)
- \( W \) = Work done (Joules, J)
- \( t \) = Time (Seconds, s)
### **Electrical Power**
1. **Using Voltage and Current:**
\[
P = VI
\]
Where:
- \( V \) = Voltage (Volts, V)
- \( I \) = Current (Amperes, A)
2. **Using Resistance:**
\[
P = I^2 R
\]
or
\[
P = \frac{V^2}{R}
\]
Where:
- \( R \) = Resistance (Ohms, Ω)
### **Mechanical Power**
1. **Using Force and Velocity:**
\[
P = F v
\]
Where:
- \( F \) = Force (Newtons, N)
- \( v \) = Velocity (m/s)
2. **Using Torque and Angular Velocity:**
\[
P = \tau \omega
\]
Where:
- \( \tau \) = Torque (Newton-meters, Nm)
- \( \omega \) = Angular velocity (radians per second, rad/s)
### **Power in Optics (Lens Power)**
\[
P = \frac{1}{f}
\]
Where:
- \( P \) = Power of the lens (Diopters, D)
- \( f \) = Focal length (Meters, m)
pls admins, when it comes to power of a lens the focal lenght is always expressed in meters .
which makes E the correct option here
now power opens = 1/focal length
1/u+1/v =1/f
for this man to see objects clearly at 25cm from the eye , the image must appear to be at 50cm
converging lenses because he IS FAR SIGHTED.
HENCE u = 25cm ( object distance) and v = -50cm (image distance formed at front of his eyes negative)
1/25+(-1/50)=1/f
(2-1)/50≈1/f
f= 50cm
power of lens =1 /f
p =1/50 dioptres
p= 0.02Dooptres
