NUMERICALS BASED ON CONVEX AND CONCAVE LENS

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Table of Contents

Numerical Problems on Spherical Lenses

Class 10 Science · Light – Reflection and Refraction

Convex Lens Practice Set

Apply the lens formula $\frac{1}{v} – \frac{1}{u} = \frac{1}{f}$ and magnification $m = \frac{v}{u}$

Ch 10 · Light

A convex lens of focal length 10 cm is placed at a distance of 12 cm from a wall. How far from the lens should an object be placed so as to form its real image on the wall?

Answer & Solution

Answer: 60 cm in front of the lens

Explanation: Since the image is real and formed on the wall, $v = +12\text{ cm}$. The focal length of a convex lens is positive, so $f = +10\text{ cm}$.

Using the lens formula:

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$$

$$\frac{1}{u} = \frac{1}{12} – \frac{1}{10} = \frac{5 – 6}{60} = -\frac{1}{60}$$

Therefore, $u = -60\text{ cm}$. The object should be placed 60 cm from the lens.

If an object of 7 cm height is placed at a distance of 12 cm from a convex lens of focal length 8 cm, find the position, nature and height of the image.

Answer & Solution

Answer: 24 cm, Real and Inverted, 14 cm

Explanation: Given $h = +7\text{ cm}$, $u = -12\text{ cm}$, and $f = +8\text{ cm}$.

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{8} + \left(-\frac{1}{12}\right) = \frac{3 – 2}{24} = \frac{1}{24}$$

So, $v = +24\text{ cm}$ (Real image behind the lens).

$$m = \frac{v}{u} = \frac{24}{-12} = -2$$

$$h’ = m \times h = -2 \times 7 = -14\text{ cm}$$

The image is real, inverted, and 14 cm tall.

An object 4 cm high is placed at a distance of 10 cm from a convex lens of focal length 20 cm. Find the position, nature and size of the image.

Answer & Solution

Answer: -20 cm, virtual & erect, h’ = 8 cm, m = 2

Explanation: Given $h = +4\text{ cm}$, $u = -10\text{ cm}$, and $f = +20\text{ cm}$.

$$\frac{1}{v} = \frac{1}{20} + \left(-\frac{1}{10}\right) = \frac{1 – 2}{20} = -\frac{1}{20}$$

So, $v = -20\text{ cm}$ (Virtual image on the same side).

$$m = \frac{v}{u} = \frac{-20}{-10} = +2$$

$$h’ = m \times h = 2 \times 4 = +8\text{ cm}$$

The image is virtual, erect, and 8 cm tall.

A small object is so placed in front of a convex lens of 5 cm focal length that a virtual image is formed at a distance of 25 cm. Find the magnification.

Answer & Solution

Answer: $m = 6$, $u = -25/6\text{ cm}$

Explanation: Given $f = +5\text{ cm}$ and $v = -25\text{ cm}$ (virtual image).

$$\frac{1}{u} = \frac{1}{v} – \frac{1}{f} = -\frac{1}{25} – \frac{1}{5} = \frac{-1 – 5}{25} = -\frac{6}{25}$$

So, $u = -\frac{25}{6}\text{ cm}$.

$$m = \frac{v}{u} = \frac{-25}{-25/6} = +6$$

Find the position and nature of the image of an object 5 cm high and 10 cm in front of a convex lens of focal length 6 cm.

Answer & Solution

Answer: 15 cm, $m = -1.5$, real & inverted

Explanation: Given $h = +5\text{ cm}$, $u = -10\text{ cm}$, and $f = +6\text{ cm}$.

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{6} – \frac{1}{10} = \frac{5 – 3}{30} = \frac{2}{30} = \frac{1}{15}$$

So, $v = +15\text{ cm}$ (Real image behind the lens).

$$m = \frac{v}{u} = \frac{15}{-10} = -1.5$$

The negative magnification indicates the image is real and inverted.

Calculate the focal length of a convex lens, which produces a virtual image at a distance of 50 cm of an object placed 20 cm in front of it.

Answer & Solution

Answer: $f = 33.34\text{ cm}$

Explanation: Given $u = -20\text{ cm}$ and $v = -50\text{ cm}$ (virtual image).

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = -\frac{1}{50} – \left(-\frac{1}{20}\right) = -\frac{2}{100} + \frac{5}{100} = \frac{3}{100}$$

$$f = \frac{100}{3} \approx +33.34\text{ cm}$$

An object is placed at a distance of 100 cm from a converging lens of focal length 40 cm. What is the nature and position of the image?

Answer & Solution

Answer: $v = 66.67\text{ cm}$ behind the lens, $m = -0.667$, real and inverted.

Explanation: Given $u = -100\text{ cm}$ and $f = +40\text{ cm}$.

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{40} – \frac{1}{100} = \frac{5 – 2}{200} = \frac{3}{200}$$

So, $v = \frac{200}{3} \approx +66.67\text{ cm}$.

$$m = \frac{v}{u} = \frac{200/3}{-100} = -\frac{2}{3} \approx -0.667$$

A convex lens produces an inverted image magnified three times of an object at a distance of 15 cm from it. Calculate focal length of the lens.

Answer & Solution

Answer: $v = +45\text{ cm}$, $f = 11.25\text{ cm}$

Explanation: An inverted image means $m$ is negative. So, $m = -3$ and $u = -15\text{ cm}$.

$$v = m \times u = -3 \times -15 = +45\text{ cm}$$

Now, calculate focal length:

$$\frac{1}{f} = \frac{1}{45} – \left(-\frac{1}{15}\right) = \frac{1}{45} + \frac{3}{45} = \frac{4}{45}$$

$$f = \frac{45}{4} = +11.25\text{ cm}$$

An object placed 4 cm in front of a converging lens produces a real image 12 cm from the lens. What is the magnification of the image? What is the focal length of the lens? Also draw the ray diagram to show the formation of the image.

Answer & Solution

Answer: $m = -3$, $f = 3\text{ cm}$

Explanation: Real image means $v$ is positive. Given $u = -4\text{ cm}$ and $v = +12\text{ cm}$.

$$m = \frac{v}{u} = \frac{12}{-4} = -3$$

$$\frac{1}{f} = \frac{1}{12} – \left(-\frac{1}{4}\right) = \frac{1}{12} + \frac{3}{12} = \frac{4}{12} = \frac{1}{3}$$

So, $f = +3\text{ cm}$. (Note: For the ray diagram, sketch an object between F and 2F forming a real, magnified image beyond 2F on the other side.)

A lens of focal length 20 cm is used to produce a ten times magnified image of a film slide on a screen. How far must the slide be placed from the lens?

Answer & Solution

Answer: $u = -22\text{ cm}$

Explanation: Since the image is formed on a screen, it must be real and inverted, so $m = -10$.

$$v = m \times u = -10u$$

Given $f = +20\text{ cm}$. Using the lens formula:

$$\frac{1}{20} = \frac{1}{-10u} – \frac{1}{u} = \frac{-1 – 10}{10u} = -\frac{11}{10u}$$

$$10u = -220 \implies u = -22\text{ cm}$$

Determine how far an object must be placed in front of a converging lens of focal length 10 cm in order to produce an erect image of linear magnification 4.

Answer & Solution

Answer: $u = -7.5\text{ cm}$

Explanation: An erect image has positive magnification, so $m = +4$.

$$v = m \times u = 4u$$

Given $f = +10\text{ cm}$. Using the lens formula:

$$\frac{1}{10} = \frac{1}{4u} – \frac{1}{u} = \frac{1 – 4}{4u} = -\frac{3}{4u}$$

$$4u = -30 \implies u = -7.5\text{ cm}$$

A convex lens of focal length 6 cm is held 4 cm from a newspaper, which has print 0.5 cm high. By calculation, determine the size and nature of the image produced.

Answer & Solution

Answer: $v = -12\text{ cm}$, $h’ = 1.5\text{ cm}$, virtual and erect

Explanation: Given $f = +6\text{ cm}$, $u = -4\text{ cm}$, and $h = 0.5\text{ cm}$.

$$\frac{1}{v} = \frac{1}{6} – \frac{1}{4} = \frac{2 – 3}{12} = -\frac{1}{12}$$

So, $v = -12\text{ cm}$ (Virtual image).

$$m = \frac{v}{u} = \frac{-12}{-4} = +3$$

$$h’ = m \times h = 3 \times 0.5 = 1.5\text{ cm}$$

A convex lens of focal length 0.10 m is used to form a magnified image of an object of height 5 mm placed at a distance of 0.08 m from the lens. Find the position, nature and size of the image.

Answer & Solution

Answer: $v = -0.4\text{ m}$, $h’ = 25\text{ mm}$, virtual & erect

Explanation: Given $f = +0.10\text{ m}$, $u = -0.08\text{ m}$, and $h = 5\text{ mm}$.

$$\frac{1}{v} = \frac{1}{0.10} – \frac{1}{0.08} = 10 – 12.5 = -2.5$$

So, $v = -\frac{1}{2.5} = -0.4\text{ m}$ (Virtual image).

$$m = \frac{v}{u} = \frac{-0.4}{-0.08} = +5$$

$$h’ = m \times h = 5 \times 5 = +25\text{ mm}$$

An erect image 2 cm high is formed 12 cm from a lens, the object being 0.5 cm high. Find the focal length of the lens.

Answer & Solution

Answer: $u = -3\text{ cm}$, $f = 4\text{ cm}$

Explanation: An erect image is virtual, so $v = -12\text{ cm}$. $h’ = +2\text{ cm}$ and $h = +0.5\text{ cm}$.

$$m = \frac{h’}{h} = \frac{2}{0.5} = +4$$

Since $m = \frac{v}{u}$, we have $4 = \frac{-12}{u} \implies u = -3\text{ cm}$.

$$\frac{1}{f} = \frac{1}{-12} – \left(-\frac{1}{3}\right) = -\frac{1}{12} + \frac{4}{12} = \frac{3}{12} = \frac{1}{4}$$

So, $f = +4\text{ cm}$.

The filament of a lamp is 80 cm from a screen and a converging lens forms an image of it on a screen, magnified three times. Find the distance of the lens from the filament and the focal length of the lens.

Answer & Solution

Answer: $u = -20\text{ cm}$, $v = +60\text{ cm}$, $f = 15\text{ cm}$

Explanation: Since the image is on a screen, it’s real and inverted ($m = -3$). The total distance $|u| + |v| = 80\text{ cm}$.

$$v = -3u \implies |v| = 3|u|$$

$$|u| + 3|u| = 80 \implies 4|u| = 80 \implies |u| = 20\text{ cm}$$

So $u = -20\text{ cm}$ and $v = +60\text{ cm}$.

$$\frac{1}{f} = \frac{1}{60} – \left(-\frac{1}{20}\right) = \frac{1}{60} + \frac{3}{60} = \frac{4}{60}$$

So, $f = 15\text{ cm}$.

An object 2 cm tall is placed on the axis of a convex lens of focal length 5 cm at a distance of 10 cm from the optical centre of the lens. Find the nature, position and size of the image formed. Which case of image formation by convex lenses is illustrated by this example?

Answer & Solution

Answer: $v = +10\text{ cm}$, $h’ = -2\text{ cm}$, real, inverted & same size. (Object at 2F)

Explanation: Note: The mathematical solution yields a different answer than previously provided keys. Given $f = +5\text{ cm}$ and $u = -10\text{ cm}$ (which is exactly $2f$).

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{5} – \frac{1}{10} = \frac{2 – 1}{10} = \frac{1}{10}$$

So, $v = +10\text{ cm}$.

$$m = \frac{v}{u} = \frac{10}{-10} = -1$$

$$h’ = m \times h = -1 \times 2 = -2\text{ cm}$$

This illustrates the case where the object is placed exactly at $2F$, forming an image at $2F$ on the opposite side, equal in size.

A converging lens of focal length 5 cm is placed at a distance of 20 cm from a screen. How far from the lens should an object be placed so as to form its real image on the screen?

Answer & Solution

Answer: $u = -6.67\text{ cm}$

Explanation: Given $f = +5\text{ cm}$ and $v = +20\text{ cm}$ (since it forms on a screen).

$$\frac{1}{u} = \frac{1}{v} – \frac{1}{f} = \frac{1}{20} – \frac{1}{5} = \frac{1 – 4}{20} = -\frac{3}{20}$$

$$u = -\frac{20}{3} \approx -6.67\text{ cm}$$

An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and nature of the image formed. Also draw the ray diagram.

Answer & Solution

Answer: $v = 16.66\text{ cm}$, $m = -0.66$, $h’ = -3.33\text{ cm}$, Real & inverted

Explanation: Given $h = +5\text{ cm}$, $u = -25\text{ cm}$, and $f = +10\text{ cm}$.

$$\frac{1}{v} = \frac{1}{10} + \left(-\frac{1}{25}\right) = \frac{5 – 2}{50} = \frac{3}{50}$$

So, $v = \frac{50}{3} \approx +16.67\text{ cm}$.

$$m = \frac{v}{u} = \frac{50/3}{-25} = -\frac{2}{3} \approx -0.667$$

$$h’ = m \times h = -\frac{2}{3} \times 5 = -\frac{10}{3} \approx -3.33\text{ cm}$$

At what distance should an object be placed from a convex lens of focal length 18 cm to obtain an image at 24 cm from it on the other side? What will be the magnification produced in this case?

Answer & Solution

Answer: $u = -72\text{ cm}$, $m = -1/3$

Explanation: “Other side” implies a real image, so $v = +24\text{ cm}$. Given $f = +18\text{ cm}$.

$$\frac{1}{u} = \frac{1}{v} – \frac{1}{f} = \frac{1}{24} – \frac{1}{18} = \frac{3 – 4}{72} = -\frac{1}{72}$$

So, $u = -72\text{ cm}$.

$$m = \frac{v}{u} = \frac{24}{-72} = -\frac{1}{3}$$

The magnification produced by a spherical lens is +2.5. What is the nature of image and lens?

Answer & Solution

Answer: Image is virtual, erect and magnified. Lens is Convex.

Explanation: The positive sign ($+$) indicates that the image is virtual and erect. The value (2.5 > 1) indicates the image is magnified. Only a convex lens can produce a virtual, magnified image.

What is the nature of the image formed by a convex lens if the magnification produced by a convex lens is +3?

Answer & Solution

Answer: Image is virtual, erect and magnified.

Explanation: A positive magnification always corresponds to an erect and virtual image. Because $m = +3$, the image is three times larger than the object.

What is the nature of the image formed by a convex lens if the magnification produced by a convex lens is –0.5?

Answer & Solution

Answer: Image is real, inverted & diminished.

Explanation: A negative magnification corresponds to a real, inverted image. Since the magnitude ($0.5$) is less than 1, the image is diminished (smaller than the object).

What is the position of image when an object is placed at a distance of 10 cm from a convex lens of focal length 10 cm?

Answer & Solution

Answer: Infinity.

Explanation: Given $u = -10\text{ cm}$ and $f = +10\text{ cm}$. Since the object is placed exactly at the principal focus ($u = -f$), the refracted rays will be parallel. Mathematically:

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} – \frac{1}{10} = 0$$

Hence, $v = \infty$.

Describe the nature of the image formed when an object is placed at a distance of 30 cm from a convex lens of focal length 15 cm.

Answer & Solution

Answer: $v = +30\text{ cm}$, real, inverted & same size.

Explanation: Given $u = -30\text{ cm}$ and $f = +15\text{ cm}$. The object is placed at $2F$ ($2 \times 15 = 30\text{ cm}$).

$$\frac{1}{v} = \frac{1}{15} – \frac{1}{30} = \frac{2 – 1}{30} = \frac{1}{30}$$

So $v = +30\text{ cm}$. Since $v = -u$, magnification is $-1$. The image is real, inverted, and the exact same size as the object.

At what distance from a converging lens of focal length 12 cm must an object be placed in order that an image of magnification 1 will be produced?

Answer & Solution

Answer: $u = -24\text{ cm}$

Explanation: For a real image of identical size, the magnification $m = -1$.

$$v = m \times u = -1 \times u = -u$$

Given $f = +12\text{ cm}$. Using the lens formula:

$$\frac{1}{12} = \frac{1}{-u} – \frac{1}{u} = -\frac{2}{u}$$

Therefore, $u = -24\text{ cm}$.

Concave Lens & Mixed Magnification Problems

Apply the lens formula considering $f$ is negative for concave lenses

Ch 10 · Light

A concave lens produces an image 20 cm from the lens of an object placed 30 cm from the lens. Calculate the focal length of the lens.

Answer & Solution

Answer: $f = -60\text{ cm}$

Explanation: A concave lens always forms a virtual image on the same side, so $v = -20\text{ cm}$. Object distance is $u = -30\text{ cm}$.

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-20} – \left(-\frac{1}{30}\right) = -\frac{3}{60} + \frac{2}{60} = -\frac{1}{60}$$

Therefore, $f = -60\text{ cm}$.

The magnification of a spherical lens is +0.5. What is the nature of lens and image?

Answer & Solution

Answer: Image is virtual, erect and diminished. Lens is Concave.

Explanation: The positive magnification indicates the image is virtual and erect. Because the magnitude ($0.5$) is less than 1, the image is diminished. Only a concave lens produces virtual, erect, and diminished images for real objects.

If an object is placed at a distance of 50 cm from a concave lens of focal length 20 cm, find the position, nature and height of the image.

Answer & Solution

Answer: $v = -14.28\text{ cm}$, Image is virtual & erect.

Explanation: Given $u = -50\text{ cm}$ and $f = -20\text{ cm}$ (since it’s a concave lens).

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-20} + \frac{1}{-50} = -\frac{5}{100} – \frac{2}{100} = -\frac{7}{100}$$

So, $v = -\frac{100}{7} \approx -14.28\text{ cm}$. The negative $v$ confirms a virtual, erect image.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image.

Answer & Solution

Answer: $v = -3\text{ cm}$, virtual & erect

Explanation: Given $u = -4\text{ cm}$ and $f = -12\text{ cm}$.

$$\frac{1}{v} = \frac{1}{-12} + \frac{1}{-4} = -\frac{1}{12} – \frac{3}{12} = -\frac{4}{12} = -\frac{1}{3}$$

So, $v = -3\text{ cm}$.

An object is placed at a distance of 50 cm from a concave lens produces a virtual image at a distance of 10 cm in front of the lens. Draw a diagram to show the formation of image. Calculate focal length of the lens and magnification produced.

Answer & Solution

Answer: $f = -12.5\text{ cm}$, $m = +0.2$

Explanation: Given $u = -50\text{ cm}$ and $v = -10\text{ cm}$.

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-10} – \left(-\frac{1}{50}\right) = -\frac{5}{50} + \frac{1}{50} = -\frac{4}{50}$$

So, $f = -\frac{50}{4} = -12.5\text{ cm}$.

$$m = \frac{v}{u} = \frac{-10}{-50} = +0.2$$

A 50 cm tall object is at a very large distance from a diverging lens. A virtual, erect and diminished image of the object is formed at a distance of 20 cm in front of the lens. How much is the focal length of the lens?

Answer & Solution

Answer: $f = -20\text{ cm}$

Explanation: A very large distance implies the object is at infinity ($u \to -\infty$). Rays from infinity converge or diverge from the principal focus. Since $v = -20\text{ cm}$, the focal length $f = -20\text{ cm}$.

A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Answer & Solution

Answer: $u = -30\text{ cm}$

Explanation: For a concave lens, $f = -15\text{ cm}$ and $v = -10\text{ cm}$.

$$\frac{1}{u} = \frac{1}{v} – \frac{1}{f} = \frac{1}{-10} – \frac{1}{-15} = -\frac{3}{30} + \frac{2}{30} = -\frac{1}{30}$$

So, $u = -30\text{ cm}$.

An object 60 cm from a lens gives a virtual image at a distance of 20 cm in front of the lens. What is the focal length of the lens? Is the lens converging or diverging? Give reasons for your answer.

Answer & Solution

Answer: $f = -30\text{ cm}$. The lens is diverging.

Explanation: Given $u = -60\text{ cm}$ and $v = -20\text{ cm}$.

$$\frac{1}{f} = \frac{1}{-20} – \left(-\frac{1}{60}\right) = -\frac{3}{60} + \frac{1}{60} = -\frac{2}{60} = -\frac{1}{30}$$

Since the focal length $f = -30\text{ cm}$ is negative, the lens must be a diverging (concave) lens.

A concave lens of 20 cm focal length forms an image 15 cm from the lens. Compute the object distance.

Answer & Solution

Answer: $u = -60\text{ cm}$

Explanation: Given $f = -20\text{ cm}$ and $v = -15\text{ cm}$.

$$\frac{1}{u} = \frac{1}{-15} – \frac{1}{-20} = -\frac{4}{60} + \frac{3}{60} = -\frac{1}{60}$$

Therefore, $u = -60\text{ cm}$.

A concave lens has focal length 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also find the magnification produced by the lens.

Answer & Solution

Answer: $u = -30\text{ cm}$, $m = 1/3$

Explanation: Given $f = -15\text{ cm}$ and $v = -10\text{ cm}$.

$$\frac{1}{u} = \frac{1}{-10} – \frac{1}{-15} = -\frac{3}{30} + \frac{2}{30} = -\frac{1}{30} \implies u = -30\text{ cm}$$

$$m = \frac{v}{u} = \frac{-10}{-30} = +\frac{1}{3}$$

Calculate the image distance for an object of height 12 mm at a distance of 0.20 m from a concave lens of focal length 0.30 m and state the nature and size of the image.

Answer & Solution

Answer: $v = -12\text{ cm}$, $h’ = 7.2\text{ mm}$, virtual and erect.

Explanation: Let’s use cm: $u = -20\text{ cm}$, $f = -30\text{ cm}$, $h = 12\text{ mm}$.

$$\frac{1}{v} = \frac{1}{-30} + \frac{1}{-20} = -\frac{2}{60} – \frac{3}{60} = -\frac{5}{60}$$

So, $v = -12\text{ cm}$.

$$m = \frac{v}{u} = \frac{-12}{-20} = 0.6$$

$$h’ = m \times h = 0.6 \times 12 = 7.2\text{ mm}$$

A concave lens has focal length of 20 cm. At what distance from the lens a 5 cm tall object be placed so that it forms an image at 15 cm from the lens? Also calculate the size of the image formed.

Answer & Solution

Answer: $u = -60\text{ cm}$, $h’ = 1.25\text{ cm}$

Explanation: Given $f = -20\text{ cm}$, $v = -15\text{ cm}$, and $h = 5\text{ cm}$.

$$\frac{1}{u} = \frac{1}{-15} – \frac{1}{-20} = -\frac{4}{60} + \frac{3}{60} = -\frac{1}{60} \implies u = -60\text{ cm}$$

$$m = \frac{v}{u} = \frac{-15}{-60} = 0.25$$

$$h’ = m \times h = 0.25 \times 5 = 1.25\text{ cm}$$

An object is placed 20 cm from (a) a converging lens and (b) a diverging lens of focal length 15 cm. Calculate the image position and magnification in each case.

Answer & Solution

Answer: (a) $v = 60\text{ cm}, m = -3$ | (b) $v = -8.57\text{ cm}, m = 0.43$

Explanation: In both cases, $u = -20\text{ cm}$.

(a) Converging ($f = +15\text{ cm}$):

$$\frac{1}{v} = \frac{1}{15} – \frac{1}{20} = \frac{4 – 3}{60} = \frac{1}{60} \implies v = 60\text{ cm}$$

$$m = \frac{60}{-20} = -3$$

(b) Diverging ($f = -15\text{ cm}$):

$$\frac{1}{v} = \frac{1}{-15} – \frac{1}{20} = -\frac{4}{60} – \frac{3}{60} = -\frac{7}{60} \implies v = -8.57\text{ cm}$$

$$m = \frac{-60/7}{-20} = \frac{3}{7} \approx 0.43$$

A 2.0 cm tall object is placed 40 cm from a diverging lens of focal length 15 cm. Find the position and size of the image.

Answer & Solution

Answer: $v = -10.91\text{ cm}$, $h’ = 0.54\text{ cm}$

Explanation: Given $h = 2.0\text{ cm}$, $u = -40\text{ cm}$, and $f = -15\text{ cm}$.

$$\frac{1}{v} = \frac{1}{-15} + \frac{1}{-40} = -\frac{8}{120} – \frac{3}{120} = -\frac{11}{120}$$

So, $v = -\frac{120}{11} \approx -10.91\text{ cm}$.

$$m = \frac{v}{u} = \frac{-120/11}{-40} = \frac{3}{11} \approx 0.27$$

$$h’ = m \times h = \frac{3}{11} \times 2.0 = \frac{6}{11} \approx 0.54\text{ cm}$$

Find the position and size of the virtual image formed when an object 2 cm tall is placed 20 cm from (a) diverging lens of focal length 40 cm and (b) converging lens of focal length 40 cm.

Answer & Solution

Answer: (a) $v = -13.33\text{ cm}, m = 0.67$ | (b) $v = -40\text{ cm}, m = 2$

Explanation: Note: The mathematical solution yields corrected magnification parameters. $u = -20\text{ cm}$.

(a) Diverging ($f = -40\text{ cm}$):

$$\frac{1}{v} = \frac{1}{-40} + \frac{1}{-20} = -\frac{3}{40} \implies v = -13.33\text{ cm}$$

$$m = \frac{-40/3}{-20} = \frac{2}{3} \approx 0.67$$

(b) Converging ($f = +40\text{ cm}$):

$$\frac{1}{v} = \frac{1}{40} + \frac{1}{-20} = -\frac{1}{40} \implies v = -40\text{ cm}$$

$$m = \frac{-40}{-20} = 2$$

The magnification produced by a spherical lens is +0.75. What is the nature of image and lens?

Answer & Solution

Answer: Image is virtual, erect & diminished. Lens is concave.

Explanation: The magnification is positive, implying a virtual and erect image. Because $m = 0.75 < 1$, the image is diminished. A concave lens exclusively forms diminished, virtual images for real objects.

The magnification produced by a spherical lens and a spherical mirror is +0.8. What is the nature of lens and mirror?

Answer & Solution

Answer: The mirror is convex and the lens is concave.

Explanation: A magnification of $+0.8$ means the image is virtual, erect, and diminished. The only mirror that forms such an image is a convex mirror, and the only lens that forms such an image is a concave lens.

The magnification produced by a spherical lens and a spherical mirror is +2.0. What is the nature of lens and mirror?

Answer & Solution

Answer: The mirror is concave and the lens is convex.

Explanation: A magnification of $+2.0$ means the image is virtual, erect, and magnified. A concave mirror forms magnified virtual images when the object is close to it. A convex lens also forms magnified virtual images when the object is placed within its focal length.

The lens A produces a magnification of –0.6 whereas lens B produces magnification of +0.6. What is the nature of lens A and B.

Answer & Solution

Answer: Lens A is Convex. Lens B is Concave.

Explanation: Lens A has a negative magnification ($-0.6$), meaning it forms a real and inverted image. Only convex lenses can form real images. Lens B has a positive but diminished magnification ($+0.6$), meaning the image is virtual and smaller. This is a property of concave lenses.

An object is 2 m from a lens which forms an erect image one-fourth (exactly) the size of the object. Determine the focal length of the lens. What type of the lens is this?

Answer & Solution

Answer: $f = -2/3\text{ m}$ (or -66.67 cm). The lens is concave.

Explanation: Given $u = -2\text{ m}$. An erect, diminished image implies a positive magnification $m = +1/4$.

$$v = m \times u = \frac{1}{4} \times (-2) = -0.5\text{ m}$$

Using the lens formula:

$$\frac{1}{f} = \frac{1}{-0.5} – \frac{1}{-2} = -2 + 0.5 = -1.5$$

$$f = -\frac{1}{1.5} = -\frac{2}{3}\text{ m}$$