determine the type and focal length of a lens that has the same power as a combination of a thin converging lens of focal length 6cm and a thin diverging lens of focal length 8cm
To determine the type and focal length of a lens that has the same power as the combination of a thin converging lens of focal length 6cm and a thin diverging lens of focal length 8cm, we can use the formula for the combined power of two lenses in contact.
The formula for the combined power (Pc) of two lenses in contact is given by:
Pc = P1 + P2
where P1 is the power of the first lens and P2 is the power of the second lens.
The power of a lens (P) is defined as:
P = 1/f
where f is the focal length of the lens.
Given that the focal length of the converging lens is 6cm, its power (P1) can be calculated as:
P1 = 1/f1 = 1/6 = 0.1667 D
Given that the focal length of the diverging lens is -8cm (negative sign indicates a diverging lens), its power (P2) can be calculated as:
P2 = 1/f2 = 1/(-8) = -0.125 D
Therefore, the combined power (Pc) of the two lenses in contact is:
Pc = P1 + P2 = 0.1667 D + (-0.125 D) = 0.0417 D
To find the focal length (f) of the lens with the same power as the combination, we can rearrange the equation for lens power:
Pc = 1/f
Substituting the calculated value for Pc, we get:
0.0417 D = 1/f
Solving for f, we find:
f = 1/0.0417 D ≈ 24 cm
Therefore, the lens that has the same power as the combination of a 6cm converging lens and an 8cm diverging lens is a converging lens with a focal length of approximately 24cm.
To determine the type and focal length of a lens that has the same power as the combination of a converging lens of focal length 6cm and a diverging lens of focal length 8cm, we can use the power formula for thin lenses.
The power of a lens (P) is given by the formula:
P = 1/f
where f is the focal length of the lens.
In this case, we have the combination of a converging lens and a diverging lens. To find the combined power of the lenses, we can add the individual powers of the lenses.
Given:
Converging lens focal length (f1) = 6cm
Diverging lens focal length (f2) = -8cm (since it's a diverging lens)
So, the power of the converging lens (P1) is:
P1 = 1/f1 = 1/6 cm^-1
And the power of the diverging lens (P2) is:
P2 = 1/f2 = 1/(-8) cm^-1
To find the combined power (P), we add the individual powers:
P = P1 + P2
P = 1/6 cm^-1 + 1/(-8) cm^-1
To simplify the expression, we find a common denominator:
P = (1*(-8) + 1*6) / (6 * (-8))
Simplifying further:
P = (-8 + 6) / (-48)
P = -2 / (-48)
P = 1 / 24 cm^-1
Now, to find the focal length (f) of the new lens, we can use the formula:
f = 1/P
Substituting the value of P:
f = 1 / (1/24)
f = 24 cm
Since the focal length of the new lens is positive (+24cm), it indicates that the new lens is a converging lens.
Therefore, the type of the lens that has the same power as the combination of a converging lens of focal length 6cm and a diverging lens of focal length 8cm is a converging lens with a focal length of +24cm.