I found an answer from spacemath.gsfc.nasa.gov

A Guide to Smartphone Sensors

Oct 30, 2018 ... apps and examine this data using a variety of mathematical operations. Smartphone apps can ... Experiment 2: Finding the Smartphone Magnetometer! ... iPhone/iPad, allows you to monitor the strength of magnetic field. Displays ... Outside house with leaves. 70. 75. 35. 59. Outside house no leaves. 76. 73.

For more information, see A Guide to Smartphone Sensors

I found an answer from goprep.co

The photograph of a house occupies an area of 1.75 cm^2 on a 35 ...

The slide is projected on to a screen, and the area of the house on the screen is ... NCERT - Physics Part-I ... Units and Measurements ... Formula: Areal magnification (m) = Aimage / Aobject , ... Since the dimension of areal magnification is L² and dimension of linear ... The radius of a cHC Verma - Concepts of Physics Part 1.

For more information, see The photograph of a house occupies an area of 1.75 cm^2 on a 35 ...

The ratio of the image's area to the object's area is known as arial magnification ( m_a ).

The ratio of the image's distance to the object's distance is known as linear magnification( m_l .

Relation between the linear and arial magnification:

The square of arial magnification is equal to linear magnification.

m_l=\sqrt{m_a}

Given that

Area of the house on the screen (image) = 1.55 m^2 = 1.55 * 10^4 cm^2

Photograph of a house Area (object) = 1.75 cm^2

Step 1: Determining the arial magnification

\text{ Arial magnification } = \frac{\text{ image's area }}{\text{ object's area }}     

\text{ Arial magnification } = \frac{1.55 * 10^4}{1.75}

\text{ Arial magnification } = 8857

Step 2: To find the linear magnification, recall the relationship between linear and arial magnification.

linear magnification m_l = \sqrt{m_a}

m_l = \sqrt{8857}

m_l = 94.1

Hence,  linear magnification of the projector-screen m_l = 94.1