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Read the following passage and answer the questions given below :
Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image.
(a) A convex lens of focal length $$\begin{aligned}& 20 \\ & \text{cm}\end{aligned}$$ is used to form an image. If an object is placed at $$\begin{aligned}& 40 \\ & \text{cm}\end{aligned}$$ from the lens, what will be the position and nature of image ?
(b) Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens.
Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image.
(a) A convex lens of focal length $$\begin{aligned}& 20 \\ & \text{cm}\end{aligned}$$ is used to form an image. If an object is placed at $$\begin{aligned}& 40 \\ & \text{cm}\end{aligned}$$ from the lens, what will be the position and nature of image ?
(b) Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens.
Show SolutionHide Solution↓
$f_1 = 30 \text{ cm} = 0.3 \text{ m, } f_2=-15\text{cm} = -0.15 \text{ m}$
$P = \frac{1}{f}$
$P_1=\frac{+1}{0.3} \text{ D} ; P_2=\frac{-1}{0.15} \text{ D}$
$\frac{1}{2}$
Equivalent power, $P = P_1+P_2$
$P = - 3.33\text{D}$
$\frac{1}{2}$
Equivalent focal length, $f=\frac{1}{P}$
$\frac{1}{2}$
$f=\frac{-1}{3.33} = - 0.3 \text{ m} = - 30 \text{ cm}$
$\frac{1}{2}$
$P = \frac{1}{f}$
$P_1=\frac{+1}{0.3} \text{ D} ; P_2=\frac{-1}{0.15} \text{ D}$
$\frac{1}{2}$
Equivalent power, $P = P_1+P_2$
$P = - 3.33\text{D}$
$\frac{1}{2}$
Equivalent focal length, $f=\frac{1}{P}$
$\frac{1}{2}$
$f=\frac{-1}{3.33} = - 0.3 \text{ m} = - 30 \text{ cm}$
$\frac{1}{2}$