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Question: A convex lens of power +6 diopter is placed in contact with a concave lens of power -4 diopter. What...

A convex lens of power +6 diopter is placed in contact with a concave lens of power -4 diopter. What will be the nature and focal length of this combination?
A). Concave, 25 cm
B). Convex, 50 cm
C). Concave, 20 cm
D). Convex, 100 cm

Explanation

Solution

Hint: In this question, we first start with the definition of power and use the expression Power of combination is equal to the sum of the power of convex lens and power of concave lens that is POWERcombination=POWERconvex+POWERconcavePOWE{R_{combination}} = POWE{R_{convex}} + POWE{R_{concave}} and get total power as PTotal=+64{P_{Total}} = + 6 - 4 diopter. It is positive. it behaves as a convex lens otherwise as a concave lens then we use the formula P=1fP = \dfrac{1}{f} to find the focal length from the power.

Formula used: POWERcombination=POWERconvex+POWERconcavePOWE{R_{combination}} = POWE{R_{convex}} + POWE{R_{concave}}, P=1fP = \dfrac{1}{f}

Complete step-by-step solution -
The ability of any lens to converge or diverge the beam of light rays is known as the power of the given lenses. The expression for power is given as the inverse of focal length f that is
P=1fP = \dfrac{1}{f} ------------------ (1)
The SI unit of power of a lens is given by Diopter which is denoted by the alphabet D. Diopter is the inverse of meter because the SI unit of the focal length is Meter.
The power of the combination is the sum of the power of all the lenses present in the combination. Therefore here the total power of the combination is the sum of the power of the concave lens that is -4 diopter and the power of the convex lens that is +6 diopter. So we can write it as
POWERcombination=POWERconvex+POWERconcavePOWE{R_{combination}} = POWE{R_{convex}} + POWE{R_{concave}}
PTotal=+64{P_{Total}} = + 6 - 4
PTotal=+2{P_{Total}} = + 2
As the total power that we are getting is +2 diopter, therefore, the combination behaves as a convex lens.
If the power is negative then the combination behaves as a concave lens.
Now for finding the focal length we use the equation (1) and substitute the PTotal=+2{P_{Total}} = + 2 in it, we will get
P=1fP = \dfrac{1}{f}
f=1P\Rightarrow f = \dfrac{1}{P}
f=1+2\Rightarrow f = \dfrac{1}{{ + 2}}
f=0.5m\Rightarrow f = 0.5m
f=0.5×100cm\Rightarrow f = 0.5 \times 100cm
f=50cm\Rightarrow f = 50cm
Hence, the focal length of the combination is 50 centimeter, therefore option B is correct.

Note: For these types of questions we need to know about how the power of lenses is calculated. We also need to know that positive power represents a convex lens that is used to correct farsightedness and if the power is negative it represents a concave lens that is used to correct nearsightedness.