Lens Power Calculation for Vision Correction
This problem requires calculating the power of a lens needed to correct vision so that a person can read a book placed at 25 cm, given their minimum distance of distinct vision is 50 cm.
Understanding the Vision Problem
- The person's near point (minimum distance of distinct vision) is 50 cm. This means they can see objects clearly only if they are at or beyond 50 cm.
- The goal is to read a book at 25 cm.
- A corrective lens is needed to make the book at 25 cm appear as if it is at 50 cm, allowing the person to see it clearly. This means the lens must form a virtual image of the object (book) at the person's near point.
Applying the Lens Formula
We use the lens formula to find the focal length ($f$). The formula is:
$ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} $- Object Distance ($u$): The distance of the book from the lens. Since the book is placed in front of the lens, $u = -25 \text{ cm}$.
- Image Distance ($v$): The distance where the virtual image should be formed for clear vision. This is the person's near point. Since it's a virtual image on the same side as the object, $v = -50 \text{ cm}$.
Calculating Focal Length
Substitute the values into the lens formula:
$ \frac{1}{f} = \frac{1}{(-50 \text{ cm})} - \frac{1}{(-25 \text{ cm})} $
$ \frac{1}{f} = -\frac{1}{50 \text{ cm}} + \frac{1}{25 \text{ cm}} $
Find a common denominator (50):
$ \frac{1}{f} = -\frac{1}{50 \text{ cm}} + \frac{2}{50 \text{ cm}} $
$ \frac{1}{f} = \frac{-1 + 2}{50 \text{ cm}} = \frac{1}{50 \text{ cm}} $
Thus, the focal length $f = 50 \text{ cm}$.
Calculating Lens Power
The power ($P$) of a lens is defined as the reciprocal of its focal length ($f$) in meters.
$ P = \frac{1}{f} $
First, convert the focal length to meters:
$ f = 50 \text{ cm} = 0.50 \text{ m} $
Now, calculate the power:
$ P = \frac{1}{0.50 \text{ m}} = 2 \text{ D} $
The required power of the lens is 2 Diopters.
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Important Questions from Optics
Consider following statements for refraction of light through prism, when angle of deviation is minimum.
A. The refracted ray inside prism becomes parallel to the base.
B. Larger angle prisms provide smaller angle of minimum deviation.
C. Angle of incidence and angle of emergence becomes equal.
D. There are always two sets of angle of incidence for which deviation will be same except at minimum deviation setting.
E. Angle of refraction becomes double of prism angle. Choose the correct answer from the options given below:
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In the figure shown below, a resistance of $150.4 \Omega$ is connected in series to an ammeter A of resistance $240 \Omega$. A shunt resistance of $10 \Omega$ is connected in parallel with the ammeter. The reading of the ammeter is __________ mA.
A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is :
